Series, Jordan–Hölder Theorem and the Extension Problem .....187 9.1 Composition Series and the Jordan–Hölder Theorem

A Course on Finite Groups Universitext

For other titles in this series, go to www.springer.com/series/223 H.E. Rose

A Course on Finite Groups H.E. Rose Department of Mathematics University of Bristol University Walk Bristol BS8 1TW, UK [email protected]

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Springer is part of Springer Science+Business Media (www.springer.com) Preface

This book is an introduction to the remarkable range and variety of finite group theory for undergraduate and beginning graduate mathematicians, and all others with an interest in the subject. My original plan was to develop the theory to the point where I could present the proofs and supporting material for some of the main results in the subject. These were to include the theorems of Lagrange, Sylow, Burn- side (Normal Complement), Jordan–Hölder, Hall and Schur–Zassenhaus amongst others, and to provide an introduction to character theory developed to the point where Burnside’s pr qs -theorem could be derived and Frobenius kernels and complements could be introduced. I have come to realise that this would have resulted in a rather long book and so some material would have to go. It was at this point that modern technology came to my aid. Solutions to the problems were also to be included, but these would have taken at least 90 rather dense pages and an appendix to this book was perhaps not the best place for this material. A number of textbooks now put solutions on a web site attached to the book which is maintained jointly by the author and the publisher. Extending this idea has allowed me to fulfil my original intentions and keep the printed text to manageable proportions. So the web site now attached to this book, which can be found by going to www.springer.com and following the product links, includes not only the Solution Appendix but also extra sections to many of the chapters and two extra web chapters. These items are listed on the contents pages, and present work that is not basic to a chapter’s topic being either slightly more specialised or slightly more challenging. Also, perhaps unfortunately, all work on character theory and applications (Chapters 13 and 14) is now on the web. As this book goes to press, about half of this web material is written and ‘latexed’, it is hoped that the remaining half will be available when the book is published or soon after. Of course, more web items could be added later. I attended Muchio Suzuki’s graduate group theory lectures given at the University of Illinois in 1974 and 1975, and so in tribute to him and the insight he gave into modern finite group theory I have ended the extended text with a discussion of his simple groups Sz(2n) as an application of the Frobenius theory.

v vi Preface

Prerequisites

This book begins with the definition of a group, and Appendices A and B give a brief résumé of the background material from Set Theory and Number Theory that is required. So in one sense, the book needs no prerequisites, only the ability to ‘think-straight’ and a desire to learn the subject. On the other hand, it would help if the reader had undertaken the following. (a) We are assuming that the reader is familiar with the material of a basic abstract algebra course, and so he or she has seen at least a few examples of groups and fields, associative and commutative operations, et cetera, and also has had some experience working in an abstract setting. (b) We are also assuming that the reader is familiar with the basics of linear algebra including facts about vector spaces, matrices and determinants, and the definitions of inner and Hermitian forms. We also use the elementary operations, similarity and rational canonical forms, and related topics. Most standard one-semester linear algebra textbooks provide more than is required. (c) It would also help if the reader had undertaken a first course on analysis which included the basic set operations, elementary properties of the standard number systems: integers Z, rational numbers Q, real numbers R, and the complex numbers C, and the standard set-theoretic methods summarised in Appendix A. (d) Lastly, some familiarity with elementary number theory would be an asset, Ap- pendix B summarises most that is required. The Euclidean Algorithm is used widely in this book, as are the basic congruence properties.

Plan of the Book

The author of an introductory group theory text has a problem: the theory is self- contained and coherent, many topics are interconnected, and several are needed more or less from the start. On the other hand, the material in a book has perforce to be presented linearly starting at Page 1. During the planning and writing of this book, I have assumed that most readers will not read it sequentially from cover to cover, but will occasionally ‘dot-about’. Hence I have allowed some ‘forward reference’, mostly for examples. The essential topics that the reader should ‘get to grips with’ first include the basic facts about groups and subgroups, homomorphisms and isomorphisms, direct products and solubility. Also some aspects of the theory of actions—conjugacy, the centraliser and the normaliser—are not far behind. Of course, as noted above, although the material has to be presented linearly, it need not be read linearly, and there are considerable advantages in presenting the basic facts of a topic— homomorphisms, for example—in one place. One consequence of this fact is that the order of the chapters has some flexibility. So Chapter 7 could be read before Chapters 5 and 6 with only a small amount of back-reference in the examples. Some group-theorists may consider it essential for students to have a good grounding in the Abelian theory before the non-Abelian theory is tackled. Similarly, Chapters 10 Preface vii and 11 can be read in either order with little back-reference required. So a possible non-linear reading of the text is Sections 2.1, 2.3, 2.4 and 4.1—the basic core of the subject, then the rest of Chapter 2, Sections 4.2, 4.3, 7.1 and 11.1 in this order, then the following sections where the reading order might be varied Part or all of Chapters 3 and 5, Sections 7.2, 7.3, and 9.1, and Chapters 6 and 10. Following this the remaining printed sections or possibly some of the web sections could be tackled. In the text, I have sometimes introduced topics ‘early’ and out of their logical order, for example, isomorphisms in Chapter 2, to deal with this point. Also, as a general rule, the ‘easier’ and/or more elementary parts of a topic come near the beginning of the chapter, and so the final sections often contain more specialised and/or challenging material.

Further Reading

The reader would do not harm studying any of the books listed in the bibliography, we suggest a few concentrating on the more recent titles. For a general further development of the finite theory try: Robinson (1982), Suzuki (1982, 1986), Aschbacher (1986), Kurzweil and Stellmacher (2004), and Isaacs (2008). Also the three volume Huppert and Blackburn (1967, 1982a, 1982b) is very com- prehensive and deals with many topics not found elsewhere. For more specialised topics, the following should be read: Doerk and Hawkes (1992) for soluble groups, Carter (1972), the ATLAS (1985), and Conway and Sloane (1993) for finite simple groups, James and Liebeck (1993), Huppert (1998) and Isaacs (2006) for character theory, and Kaplansky (1969), Fuchs (1970, 1973), and Rotman (1994) for infinite Abelian groups. Of course, some of the older books still have much to offer, these include Burnside (1911, reprinted 2004), Kurosh (1955), Scott (1964) and Rose (1978)—no relation! Although 45 years old, in my opinion, Scott’s book remains one of the best introductions to the subject. viii Preface

All errors and omissions that are still present in the text and/or web pages are entirely my fault, please contact me with details at [email protected] General comments, including comments on the correctness and/or clarity of the text, or shorter, clearer or better solutions to the problems (which could be added to the web site), are also welcome.

Acknowledgements I have received a considerable amount of assistance during the writing of this book for which I am extremely grateful. First, from my family (especially from my wife Rita), from colleagues, both academic and computational (especially Richard Lewis and Peter Burton), at the University of Bristol, and from the staff (both editorial and ‘Latex’ specialist) at Springer Verlag. I have given courses based on preliminary versions of this book many times in Bristol, my students have helped me to clarify many points and I thank them for this. But my main debt of gratitude goes to those who read a ﬁnal draft of the text and cleared up many inconsistencies and errors on my part. These include the referees appointed by Springer Verlag, John Bowers (formally of the University of Leeds), Robin Chap- man (Universities of Bristol and Exeter), Robert Curtis for Chapter 12 (University of Birmingham), and Ben Fairbairn (University of Birmingham). These last four spent many hours going through the manuscript, and improved it greatly—they are forever in my debt. Bristol Harvey Rose Contents

1 Introduction—The Group Concept ...... 1

2 Elementary Group Properties ...... 11 2.1BasicDeﬁnitions...... 11 2.2Examples...... 17 2.3 Subgroups, Cosets and Lagrange’s Theorem ...... 24 2.4 Normal Subgroups ...... 30 2.5Problems...... 34

3 Group Construction and Representation ...... 41 3.1Permutations...... 42 3.2 Permutation Groups ...... 48 3.3 Matrix Groups ...... 52 3.4GroupPresentation...... 55 3.5Problems...... 59

4 Homomorphisms ...... 67 4.1HomomorphismsandIsomorphisms...... 69 4.2 Isomorphism Theorems ...... 74 4.3 Cyclic Groups ...... 79 4.4 Automorphism Groups ...... 81 4.5Problems...... 84

5 Action and the Orbit–Stabiliser Theorem ...... 91 5.1Actions...... 92 5.2ThreeImportantExamples...... 99 5.3Problems...... 107

6 p-Groups and Sylow Theory ...... 113 6.1 Finite p-Groups ...... 114 6.2 Sylow Theory ...... 119 6.3Applications...... 126 6.4Problems...... 131 ix x Contents

7 Products and Abelian Groups ...... 139 7.1 Direct Products ...... 140 7.2 Finite Abelian Groups ...... 146 7.3 Semi-direct Products ...... 151 7.4Problems...... 159 8 Groups of Order 24, Three Examples ...... 165 8.1 Symmetric Group S4 ...... 165 8.2 Special Linear Group SL2(3) ...... 172 8.3 Exceptional Group E ...... 177 8.4Problems...... 183 9 Series, Jordan–Hölder Theorem and the Extension Problem .....187 9.1 Composition Series and the Jordan–Hölder Theorem ...... 188 9.2ExtensionProblem...... 196 9.3Problems...... 205

10 Nilpotency ...... 209 10.1 Nilpotent Groups ...... 210 10.2 Frattini and Fitting Subgroups ...... 217 10.3Problems...... 223 11 Solubility ...... 229 11.1 Soluble Groups ...... 230 11.2 Hall’s Theorems and Solubility Conditions ...... 236 11.3Problems...... 243 12 Simple Groups of Order Less than 10000 ...... 249 12.1SteinerSystems...... 250 12.2 Linear Groups ...... 254 12.3 Unitary Groups ...... 265 12.4 Mathieu Groups ...... 267 12.5Problems...... 270 Appendices A to E ...... 277 A Set Theory ...... 277 B Number Theory ...... 284 C Data on Groups of Order 24 ...... 289 D Numbers of Groups with Order up to 520 ...... 293 E Representations of L2(q) with Order < 10000 ...... 295 Bibliography ...... 297 Notation Index ...... 301 1—SymbolIndex...... 301 2—Notation for Classes of Groups ...... 303 3—Notation for Individual Groups ...... 304 Index ...... 305 Web Contents

3 Group Construction and Representation ...... 321 3.6 Representations of A5 3.7 Further Problems

4 Homomorphisms ...... 331 4.6 The Transfer 4.7 Group Presentation, Part 2 4.8 Further Problems

5 Action and the Orbit–Stabiliser Theorem ...... 351 5.4 Transitive and Primitive Permutation Groups 5.5 Further Problems

6 p-Groups and Sylow Theory ...... 361 6.5 Applications 2—Burnside’s Normal Complement Theorem and Groups with Cyclic Sylow Subgroups 6.6 Further Problems

7 Products and Abelian Groups ...... 371 7.5 Inﬁnite Abelian Groups—A Brief Introduction

9 Series, Jordan–Hölder Theorem and the Extension Problem .....379 9.4 Schur–Zassenhaus Theorem 9.5 Further Problems

12 Simple Groups of Order Less than 10000 ...... 387 12.6 Simple Groups of Order Less than 1000000, Iwasawa’s Lemma and a Method for generating Steiner Systems for some Mathieu Groups 12.7 Further Problems

13 Representation and Character Theory ...... 401 13.1 Representations and Modules

xi xii Web Contents

13.2 Theorems of Schur and Maschke 13.3 Characters and Orthogonality Relations 13.4 Lifts and Normal Subgroups 13.5 Problems

14 Character Tables and Theorems of Burnside and Frobenius .....433 14.1 Character Tables 14.2 Burnside’s pr qs -Theorem 14.3 Frobenius Groups 14.4 Problems

Solution Appendix—Answers and Solutions, Problems 2 to 12, A and B ...... 461 Problem2 ...... 461 Problem3 ...... 472 Problem4 ...... 481 Problem5 ...... 489 Problem6 ...... 497 Problem7 ...... 508 Problem8 ...... 515 Problem9 ...... 521 Problem10...... 524 Problem11...... 532 Problem12...... 538 ProblemA...... 544 ProblemB...... 545 Chapter 1 Introduction—The Group Concept

Groups are all-pervasive in mathematics, there is hardly a branch of the subject that does not use them in one way or another. They are also widely used in many branches of the physical sciences. In one sense, this is to be expected because groups are quite often formed when an operation like multiplication or composition is applied to a set or system. Groups occur as number systems or collections of matrices, in permutation theory, as the symmetries of geometrical objects or as sets of maps, and in many other guises. Also the theory contains many elegant, dramatic and illu- minating theorems. Group theory has developed sometimes slowly but at other times by great leaps and bounds over the past two centuries. Often ideas and results appeared first im- plicitly before they were explicitly written down. For example, it is thought that Galois, in the 1820s, was the first to write down the axioms of a group, but some forty years earlier Lagrange was working with permutations of the roots of equations and proved a result which led to the famous theorem that now bears his name, the comments on actions given in the Introduction to Chapter 5 also apply here. Galois introduced a number of other basic notions including, for example, simple groups and normal subgroups. In 1850, Cayley showed that every group can be represented as a permutation group, and much of the nineteenth century work dealt with this aspect of the theory. Results started to appear more quickly, Sylow produced his ground-breaking work on p-subgroups in 1872, characters and representation theory were introduced around the turn of the century, and Hall’s extensions of Sy- low’s work appeared in the 1920s and 1930s—to mention only a few of the many major developments. Another surge began around 1950 and led in the early 1980s to the completion of the classification problem for finite simple group, hereafter referred to as CFSG, which must surely rank amongst the greatest achievements in all mathematics. A number of important corollaries have followed from this work, for example, the positive solution of the Restricted Burnside Problem (page 27). The purpose of this book is to introduce the reader to the fine branch of mathematics called group theory—there is a ‘great story’ to tell, and we hope that it will encourage you, the reader, to develop an abiding interest in the subject, and a desire to look further and deeper into the theory.

H.E. Rose, A Course on Finite Groups, 1 Universitext, DOI 10.1007/978-1-84882-889-6_1, © Springer-Verlag London Limited 2009 2 1 Introduction—The Group Concept

Group Examples

A group is a mathematical system (or set) with a single operation. We begin by considering two familiar examples; the ﬁrst is the integers Z. The elements are

...,−2, −1, 0, 1, 2, 3,..., and the operation is standard addition ‘+’. There are a number of basic axioms from which almost all additive properties follow. The ﬁrst and in some ways the most important is closure, or being well-deﬁned; that is,

if a,b ∈ Z, then a + b ∈ Z.

This is equivalent to stating that + is an operation (Appendix A). Some so-called partial systems have been studied, but all systems considered in this book satisfy an axiom of this type. The next property is associativity; that is,

for all a,b,c ∈ Z we have (a + b) + c = a + (b + c).

When forming this sequence of additions, we obtain the same answer if we ﬁrst add a to b, and then add the result of this addition to c,orifweﬁrstaddb to c and then form the sum of a and the result of this last addition. Some algebraic systems lack this property but, in general, non-associative systems have limited uses unless some more complex rule is applied—for example, in Lie algebras—and again we shall not consider such systems in this book. In Z, a natural question to ask is: Does the equation

a + x = b (1.1) have a solution x? In ancient times mathematicians only ‘allowed’ this equation to have a solution if b>a,thatis,ifx is positive. But this is very restrictive, and in the group Z Equation (1.1) is always uniquely soluble, and so we need to introduce the ‘zero’ and ‘negative’ integers. The zero 0 satisﬁes

a + 0 = 0 + a = a for all a ∈ Z; in the sequel, we use the term neutral element for the entity 0; see the discussion on page 4. Further, we introduce the negative integers by

for all a ∈ Z there exists a unique c ∈ Z that satisﬁes a + c = 0.

We usually write −a for c (and a−1 for c if we are using a multiplicative notation as is the case for almost all groups discussed in this book), and we call it the inverse of a. It is now easy to show that (1.1) always has a unique solution. The system Z has one extra basic property not shared by all groups: it is commutative,orAbelian. This is given by

for all a,b ∈ Z we have a + b = b + a, 1 Introduction—The Group Concept 3 the result of several additions does not depend on the order of the terms in the sum. To recap, Z has the four basic properties: Closure, associativity, a neutral element and inverses, and it has the extra property of Abelianness. Note also it is a countably infinite system. For our second example, we consider another familiar system—symmetries of an equilateral triangle. Groups of symmetries provide a good range of examples, they are widely used in both mathematical and physical systems, for example, by chemists when they are studying the crystal structure of matter. The group of symmetries of a triangle has two aspects which are different from those in our first example: It is finite, and it is not Abelian, but as we shall see below it shares the four basic properties with Z. Consider an equilateral triangle with vertices labelled A,B,C. This geometric object has a number of symmetries, that is, transformations (rotations and reflections) that give another copy of the original triangle. We work in the standard Euclidean plane.

A C B α α → −→ −→ → C B B A A C ||| ↓↓↓βββ

A C B α2 α2 → −→ −→ → B C A B C A

The elements of the group are the six rotations and reflections of the triangle illustrated above, and the operation is composition; that is, do one rotation or reflection, and then do another on the result of the first. We take the basic rotation α to be clockwise about the centre of the triangle by the angle 2π/3,seethetoprowofthe diagram above. Note that three applications of α (that is, α3, a rotation by 2π) maps the triangle to itself identically, and so can be taken as the neutral element. This also shows that the inverse of α is α2; see the bottom row in the diagram. The triangle has three reflections, they are mirror transformations about a line through a vertex and the centre of the corresponding opposite edge. One, labelled β (three times) in the diagram, is about a vertical line through the top vertex of the corresponding triangle and the centre of its base. The other two reflections can be generated as follows. If we first apply α to the top-left triangle and then apply β to the result, we obtain the middle triangle in the bottom row. This gives the second reflection of the top-left triangle, now about a line through the bottom right-hand vertex B and the centre of the opposite edge AC. Note that we would obtain the same result if we first applied β to the top-left triangle, and then applied α2 to the result. We can obtain the third reflection if we repeat this construction but begin by apply- 4 1 Introduction—The Group Concept ing α2 instead of α to the top-left triangle. Incidentally, this shows that the group is not Abelian (that is, not commutative as βα = α2β = αβ). The inverse of a reflection is itself: Two applications of a reflection gives the neutral element. We call a group element of this type an involution, and we shall see later that these elements play an important role in the theory. It is straightforward to check that this system is closed and has the associativity property (reader, try a few cases). Hence the system contains the six symmetries of the triangle: Neutral element (where a triangle is mapped to itself identically), α, α2, β, αβ, and α2β, and it satisfies the four basic group axioms (closure, associativity, and possession of a neutral element and inverses for all of its elements) as in our first example. It is also finite and not Abelian. Later we shall call this system the dihedral group of the triangle and denote it by D3.

Abstract Groups and Representations

With these and other examples in mind, we define a group as a system with a single operation satisfying the four basic properties (axioms) described above, the formal definition is given at the beginning of Chapter 2. Also, Section 2.2 provides a substantial list of examples. The following point is important. In both cases, the examples given above are particular ‘instances’ of the group in question; we call them representations of the group. Referring to the second example, another representation is given by considering the set of permutations of the set {1, 2, 3} with composition as the operation; see page 19 and Section 3.1. A third representation using 2 × 2 matrices is given in Problem 4.2. Each group example given in this book is a representation of its corresponding abstract group, see the discussion in the Introduction to Chapter 3. Right from the start this is a characteristic feature that the reader should note, and we use a corresponding nomenclature. So when discussing a group in the abstract we call it a ‘group’, but when discussing a particular example or representation, say using permutations or matrices or some analogous construction, we call it a ‘permutation group’ or a ‘matrix group’ or analogous group. Similarly, the ‘neutral element’ which we always denote by e is the element in the abstract group satisfying the relevant axiom, and it has many instances or representations in particular groups. In the examples given in Section 2.2,itis0,or1,or−3, or a matrix, or a collection of maps, or the point at infinity on a curve, et cetera. The term ‘neutral element’ is not standard, nor is it ‘new’; for example, it occurs in Cohn (1965), page 50. Some authors use either ‘identity’ or ‘trivial element’; the first only really applies to an ‘operation of multiplicative type’, whilst the second gives the wrong impression—this element is an important one in the group, and certainly not trivial in the usual sense of this word. We have extended this nomenclature to the single element (sub)group e which we call the neutral (sub)group, see Definition 2.12.Alsoan ‘inverse’ can be an additive inverse, or a multiplicative inverse, or an inverse matrix, et cetera. 1 Introduction—The Group Concept 5

Classes of Groups

The class of all groups is a large one. Set-theorists call it a proper class as opposed to a set, but as we are taking the usual naive view of set theory (Appendix A) we shall treat sets and classes synonymously. We shall see that it is convenient to consider subclasses defined by some of the basic properties. For example, groups can be finite or infinite, and Abelian or non-Abelian; these distinctions are fundamental. We shall study further distinctions later, for instance, in Chapter 11 between ‘soluble’ and ‘non-soluble’ groups. Here we divide the class of all groups into four subclasses and, as we shall see, both the theory and the actual groups in each subclass have distinct characteristics.

The ﬁrst subclass contains the

Finite Abelian Groups In Chapter 7, we show that groups in this class can be characterised completely, and they have a particularly simple form—that is, as ‘products of cyclic groups’. So in one sense they are ‘a bit boring’; but in an application, if we know apriorithat the group or groups under discussion are in this subclass, then we can be sure that they take this simple form which can have a major inﬂuence on the result. A good example occurs in the theory of rational points on elliptic curves discussed on page 22. Amongst our four subclasses, this is the only one for which we have a complete description of all of the groups involved.

The second subclass contains the

Finite Non-Abelian Groups Most of the work in this book deals with groups in this class. For finite groups in general, there is a strong interplay between the ‘group theory’ and the ‘number theory’ of the group in question. In part this is a consequence of Lagrange’s Theorem which states that the order (number of elements) of a subgroup H of a group G must divide the order of G; so the prime factorisation of the order of a finite group is an important invariant of the group. One major development from this is the Sylow theory discussed in Chapter 6 which asserts the existence of subgroups with prime power order. Another important distinction in the theory is between the so-called ‘simple’ and ‘non-simple’ groups; see the definition on page 33. The Jordan–Hölder Theorem states, roughly speaking, that all finite (and some infinite) groups can be ‘built up’ from simple groups using ‘extensions’; this will be discussed in Chapter 9—note that one of the main aims our work is to describe all groups. A theory of extensions has been developed, but a considerable amount of work and many new ideas will be needed before it can be described as finished; see Section 9.2. On the other hand, a complete list of finite simple groups is now known, much of the development work was undertaken between 1955 and 1985 and, as noted above, it forms one of the crowning achievements of twentieth century mathematics. We give a brief introduction to this topic in Chapter 12. Hence considerable progress has been made in the theory of finite non-Abelian groups and this will be discussed in the following chapters and web appendices, but work still needs to be done. 6 1 Introduction—The Group Concept

The third subclass contains the

Infinite Abelian Groups For infinite groups in general, number theory only plays a small role, but questions concerning cardinality can be important. The so-called finitely-generated Abelian groups are similar to those in the first subclass as they can be represented as products of cyclic groups. But many groups in this class are not finitely generated, for example, the rational numbers with addition or the positive reals with multiplication. Apart from a brief survey in Web Section 7.5 we shall not deal with these groups in this book. Much of the work develops ideas from linear algebra, and good introductions to this topic are given in Kaplansky (1969), and Fuchs (1970, 1973).

The ﬁnal subclass contains the

Infinite Non-Abelian Groups This is perhaps the least well-understood part of the theory. A number of extensions of the finite theory have been studied, but no general classification is known, and it seems unlikely that one will be found in the near future. One approach is to use topology. For example, the reals have a natural (metric) topology, and the interplay between the group theory and the topology of this system can be exploited to gain new insights. Since 1950 a number of long- standing problems have been solved, often showing that these groups are more complicated than previously thought; for example, see Problem 6.7. A good introduction is given in Kurosh (1955), also Robinson (1982) discusses a number of aspects of this part of the theory. Infinite groups with some kind of ‘finiteness condition’, such as being ‘finitely generated’ or ‘finitely presented’, have also been widely studied.

Summary of the Book

Below we give a brief summary of the contents of the printed Chapters 2 to 12, Appendices A to E,theWeb Chapters 13 and 14, and the Web Appendices.

Chapter 2—Elementary Group Properties The basic entities—semigroups, groups, subgroups, cosets, normal subgroups and simple groups—are deﬁned, La- grange’s Theorem is derived, and the second section lists a number of standard examples.

Chapter 3—Group Construction and Representation The main construction methods and group representations are discussed. Firstly, permutations are introduced, the symmetric and alternating groups are deﬁned, and an elementary proof of the simplicity of An for n>4 is given. Secondly, matrix groups are brieﬂy considered, and lastly group presentations are introduced (this topic is completed in Web Section 4.7 once the First Isomorphism Theorem has been proved). Web Sec- tion 3.6 discusses some of the various representations of the alternating group A5 to illustrate the fact that groups can have a wide range of representations. 1 Introduction—The Group Concept 7

Chapter 4—Homomorphisms The natural maps, called Homomorphisms (and Isomorphisms when the map is a bijection), and factor groups are introduced, and the four fundamental Isomorphism and Correspondence Theorems are derived. Cyclic groups and the basic properties of the automorphism group of a group are described. There are two Web Sections 4.6 and 4.7. The ﬁrst introduces the ‘transfer’ which provides a useful example of a ‘real’ homomorphism, and the second completes the work on group presentations begun in Chapter 3.

Chapter 5—Action and the Orbit-stabiliser Theorem This chapter, the last giving the basic material, introduces ‘actions’ which bring together a number of useful constructions, and gives a proof of the Orbit-stabiliser Theorem. It also describes three important particular actions: the coset action, the conjugate element action leading to centralisers and the Class Equations, and the conjugate subgroup action leading to normalisers and the N/C-theorem. Web Section 5.5 extends the work on permutation theory begun in Chapter 3, and includes a discussion of ‘transitive’ and ‘primitive’ permutation groups, and Iwasawa’s simplicity lemma.

Chapter 6—p-Groups and Sylow Theory The basic theory of p-groups (where all elements have order a power of p) is discussed, and the five Sylow theorems are derived—these results form one of the most important aspects of the finite theory. There are then two sections of applications, the first gives (a) some facts about groups whose orders have a small number of factors, (b) proves the so-called Frattini Argument, and (c) introduces nilpotent groups. The second is Web Section 6.5 which gives some more substantial applications including a proof of Burnside’s Nor- mal Complement Theorem and a discussion of groups all of whose Sylow subgroups are cyclic.

Chapter 7—Products and Abelian Groups Direct products are introduced, and two proofs of the Fundamental Theorem of Abelian Groups are presented; see page 5. The third section discusses ‘semi-direct products’, a variant of the direct product construction, and the groups of order 12 are described (they can all be treated as semi-direct products). Some basic facts, but no proofs, concerning infinite Abelian groups are given in Web Section 7.5. Except for some problems, this is the only point where specifically infinite groups are considered in any detail.

Chapter 8—Groups of Order 24, Three Examples No new theory is presented in this chapter, but three groups of order 24 are discussed in some detail. The work constructs their subgroups including those of Frattini and Fitting, the subgroup lattice, series and some of their representations. Appendix C, see pages 289 to 292, gives data on the remaining twelve groups of order 24. The purpose of this chapter is to challenge the reader to think more about the objects he or she is studying, and to ask questions. For example: can the centre of a group equal its derived subgroup or its Frattini subgroup? This chapter is also intended to motivate the remaining topics, that is series, simple groups and (on the web) representation theory. 8 1 Introduction—The Group Concept

Chapters 9—Series, Jordan–Hölder Theorem and the Extension Problem This is the ﬁrst of three shorter chapters dealing with series and the normal subgroup structure of groups. In the ﬁrst of these, we prove the theorem of Jordan and Hölder on composition series—this demonstrates the importance of simple groups to the theory. Secondly, we present a brief introduction to extension theory—that is the construction of complex groups using some of their subgroups as components, and we discuss one substantial example.

Chapter 10—Nilpotency Nilpotent groups lie between Abelian and soluble groups, and this second shorter chapter continues the work on these groups begun earlier. There is a surprising number of equivalent deﬁnitions which shows the importance of the notion. The second section discusses two ‘special’ subgroups of a group—the Frattini and Fitting subgroups; they have some remarkable properties (including being nilpotent), and extensions of the latter have proved useful in the completion of CFSG.

Chapter 11—Solubility After a brief historical introduction the last of the shorter chapters introduces the basic facts about soluble groups, and discusses a number of equivalent conditions. The most important is due to Philip Hall and extends the Sylow theory in the soluble case.

Chapter 12—Simple Groups of Order Less than 10000 This is another ‘descriptive’ chapter giving an account of simple groups with order less than 10000. We introduce Steiner systems—their automorphisms provide a new way to construct groups, prove the simplicity of the linear (matrix) groups Ln(q), and discuss one ‘classical’ (U3(3),aunitary group) and one ‘sporadic’ (M11, the ﬁrst Mathieu group) group in detail. Some numerical data is also given but many proofs are omitted. Appendix E, see page 295, gives data on the groups L2(q), and an appendix at Web Section 12.6 provides more information about Steiner systems for Mathieu groups, and data on simple groups of order less than 106.

Appendix A—Set Theory and Appendix B—Number Theory These two appendices give the basic deﬁnitions and results for the work on sets and number theory which underlie all the material in the book.

Appendices C, D and E These appendices provide data on several aspects of the theory. The ﬁrst, C, lists properties of groups of order 24 (this is an appendix to Chapter 8), D details the number of groups with order up to 520, and E provides some representations of the linear groups L2(q) (this is an appendix to Chapter 12).

Web Chapter 13—Representation and Characters A brief introduction to representation and character theory is presented sufﬁcient for the applications given in Web Chapter 14. This theory includes the basic deﬁnitions, Schur’s Lemma and Maschke’s theorem, the orthogonality relations, and ‘lifts’, et cetera. This chapter is entirely theoretical except for the examples. 1 Introduction—The Group Concept 9

Web Chapter 14—Character Tables, and Theorems of Burnside and Frobe- nius We give three applications of the work in Web Chapter 13 which have strong connections with the earlier material. First, we construct some character tables including those for most of the groups of order 12 or less, and some others including several of order 24 discussed in Chapter 8. These character tables provide a surprising amount of information concerning the groups in question. Second, we prove Burnside’s pr qs -theorem (this completes the proof of Hall’s Theorem given in Chapter 11). The third section introduces the Frobenius ‘kernel’ and ‘complement’, gives a proof of Frobenius’s Theorem concerning these notions, and ﬁnally it provides some applications of this theorem including a discussion of Suzuki groups.

Web Solution Appendix This includes answers, hints on solutions, and in some cases full solutions, for all of the problems given in Chapters 2 to 12, and Appendices A and B.

Developing the theory and proving results are of course important, but two other aspects are also important.

Problems Each chapter ends with a sequence of problems for the reader to try of varying difficulty partly as indicated with a star suggesting a greater challenge. Some readers may find it difficult to decide which problems start with and which are the most important, so some of these have been marked with the symbol . These are all fairly straightforward, theoretical, and contain minor results that are used in the main part of the text. Other problems ask for examples to be constructed, these have no indication mark but should also be tackled early. As noted above, hints, sketch solutions, or in some cases detailed treatments of problems, are given in Web Solution Appendix on the web site attached to this book.

‘Actual groups’ We are studying groups, and so it seems essential to us that the reader ‘sees’ and ‘experiences’ as many ‘actual’ or ‘concrete’ groups as possible. This will, we hope, illuminate the theory and so induce a greater understanding in general. Some parts of the text and a number of the problems are given over to this aspect including the whole of Chapter 8.

Computers in Group Theory

During the past thirty years, and more so recently, computers have become an in- creasingly useful tool in pure mathematics, as well as in most other branches of mathematics, many branches of the physical sciences, and beyond. In group theory, they are particularly useful for doing matrix and permutation calculations, and for producing examples. But they can also be used for more sophisticated constructions, for example, looking for subgroups of a group or constructing homomorphisms. A number of computer algebra packages, some ‘free’ and some commercially available, have been developed over the past decade or so, and the reader is encouraged 10 1 Introduction—The Group Concept to make use of at least one of these while reading this book. Also a number of the problems are best tackled using one of these packages. While writing this book, we have made extensive use of the computer algebra package called GAP—Groups, Algorithms, and Programming. This package has many authors based in Aachen in Germany, St. Andrews in Scotland, and at many other sites; we would like to take this opportunity to compliment these authors on the excellence of their product. It is available free from the St. Andrews web site at

http://www-gap.dcs.st-and.ac.uk/~gap

We have also made some use of the commercially available package called MAGMA which incorporates many aspects of the GAP program. One point should be borne in mind whilst working with any of these packages, and it is one that we emphasise several times in this book. In a particular calculation, the program can only deal with a specified representation of the group under discussion, say as a permutation group or as a matrix group. The package GAP is particularly good when working with permutation groups, but it also deals well with matrix groups defined over a specific field and with presentations. Chapter 2 Elementary Group Properties

In this chapter, we introduce our main objects of study—groups. A general overview including some historical comments was given in Chapter 1. More detail on the history of the theory can be found in Wussing (1984), van der Waerden (1985), and at www-gap.dcs.st-andrews.ac.uk/~history/. Here we give the basic deﬁnitions and an extensive list of examples, introduce subgroups and cosets, normal subgroups and simple groups, and prove the ﬁrst major result in the theory—Lagrange’s The- orem.

2.1 Basic Deﬁnitions

We begin by deﬁning the group concept. Maps between groups will be discussed in Chapter 4. As a preliminary to this we introduce semigroups as follows.

Definition 2.1 A semigroup is a non-empty set X ={...,x,y,z,...} together with a binary operation (page 281) which satisfies the following two conditions (axioms): (i) it is closed,orwell-defined: for all x,y ∈ X, we can perform the operation x y and

x y ∈ X,

(ii) it is associative: for all x,y,z ∈ X,

x (y z) = (x y) z.

Note that (i) is implied by the deﬁnition of the operation ; see the comments below Deﬁnition 2.2. H.E. Rose, A Course on Finite Groups, 11 Universitext, DOI 10.1007/978-1-84882-889-6_2, © Springer-Verlag London Limited 2009 12 2 Elementary Group Properties

Examples The following sets with operations are semigroups. (a) The positive integers with addition. (b) The set of all one-variable functions with domain and codomain R, and with the operation of composition of functions.

There is an extensive theory of semigroups which is of particular interest in some branches of analysis and combinatorics. Also a number of similar systems that are not quite groups have been studied, for instance, the operation may be only partially deﬁned, or there may be a neutral element but no inverses, et cetera. We shall not consider these systems; Bruck (1966) provides a good introduction.

Definition 2.2 A group (G, ) is a semigroup which satisfies the following extra conditions (axioms): (iii) G contains a unique element e which satisfies, for all g ∈ G,

e g = g e = g,

(iv) for each g ∈ G, there exists a unique g ∈ G satisfying

g g = g g = e.

The element e is called the neutral element of the group G, see page 4. Some authors use the term identity for e, and if the operation () is written additively (+) then it is called the zero and denoted by 0. There are a number of redundancies in this deﬁnition—in particular, in axioms (i), (iii) and (iv). Strictly speaking, (i) is un- necessary as it is implied by the fact that is an operation; see Appendix A.Butwe have left it in to remind the reader that closure is vitally important—this property must be checked whenever it is required to show that a particular set and product form a group. For (iii) and (iv), see Theorem 2.5.

Deﬁnition 2.3 A group (G, ) is called Abelian, or occasionally commutative,if its operation is commutative: For all g,h ∈ G

g h = h g.

The term ‘Abelian’ commemorates the Norwegian mathematician Niels Abel who died at the age of 27 in 1829. He was working on solutions to polynomial equations, and needed to apply a condition similar to the one above; see the Introduction to Chapter 11 and van der Waerden (1985), page 88.

Examples We give four here, and an extensive list in the next section. (a) The set {1, −1} with the operation of standard multiplication forms a finite Abelian group which we denote by T1 (one copy of ‘two’). The neutral element is 1, and each element is self-inverse. (b) The set of permutations of a set {1, 2, 3}. The elements are the six permutations of this set, and the operation is composition: Do the first permutation, then do the 2.1 Basic Definitions 13

second permutation on the result of the first. For example, if the first permutation maps 1 → 1, 2 → 3 and 3 → 2, and the second maps 1 → 3, 2 → 2 and 3 → 1, then their composition maps 1 → 3, 2 → 1 and 3 → 2. The neutral element is the permutation that moves no symbols, and the inverse of a permutation is its reverse (Section 3.1). This system forms a finite non-Abelian group which we denote by S3 and call the symmetric group of the set {1, 2, 3}. Reader, why is this group not Abelian? (c) The positive real numbers R+ with multiplication form an infinite Abelian group. The neutral element is 1, and the inverse of x is 1/x. (d) The set of all non-singular 2 × 2 matrices having rational number entries with the operation of matrix multiplication is an example of an infinite non-Abelian group, it is denoted by GL2(Q) and called the 2×2 general linear group over Q. The neutral element is I2, the 2-dimensional identity matrix, and inverses exist by definition.

The symbols (G, ), G, H , J ,orK, sometimes with primes or subscripts, will always denote groups. We use lower case Roman letters a, b, c, d, g, h, j, k, and l, again sometimes with primes or suffixes, to stand for group elements, and we use x, y and z for set elements or occasionally for group elements following the usual mathematical convention that these letters denote entities which satisfy a proposition or equation. The words ‘operation’, ‘multiplication’ and ‘product’ are used more or less synonymously: If g,h ∈ G we say that we apply the operation to form the product g h, or we multiply g by h to obtain g h. The underlying set of a group G is the set of elements of G stripped of its operation; where there is no confusion, this will also be denoted by G. Also, we sometimes say that a group G is generated by a set X,orX is a generating set for G, where X is a subset of the underlying set of G, and we write G =X. This means that the collection of all products of powers (both positive and negative) of elements of X coincides with G. For example, the set {1} is a generating set for Z, that is, Z =1 because every integer can be expressed as a sum of 1s or −1s. Note that a group may have many different generating sets, and it always has at least one because the underlying set of G clearly acts as a generating set for G. This notion is defined formally in Definition 2.16. We also write e for the (unique) group containing the single element e (with operation e e = e), we call it the neutral group. Some authors use the term ‘trivial group’ for e; it is an important component of a group and certainly not ‘trivial’ using the normal meaning of this word, hence we shall not use this term; see also the comments on page 4. We noted above that Definition 2.2 can be weakened considerably without af- fecting our objects of study. Consider

Definition 2.2 A group (G, ) is a semigroup, see Definition 2.1, which satisfies the following two conditions: (iii) there is an element f ∈ G with the property: f g = g, for all g ∈ G; (iv) for each g ∈ G, and with f as in (iii), there exists h ∈ G satisfying h g = f . 14 2 Elementary Group Properties

These conditions imply that G has at least one left neutral element f , and each g ∈ G has at least one left inverse h relative to f . Definition 2.2 is equivalent to Definition 2.2; see also Problem 2.4. Note that this equivalence is useful, for when checking if the group axioms hold for a particular set and map, once closure and associativity have been established (Axioms (i) and (ii)), it is not necessary to prove that either the neutral element or the inverse operation is unique, or two-sided, because these properties follow by Theorem 2.5 below. Also, if we find an inverse of an element g, then we can be sure that it is the unique inverse of g, again by Theorem 2.5. We begin with the following result: In all groups, the only element which equals its square is the neutral element (in algebra generally, such elements are called idem- potents).

Lemma 2.4 Let (G, ) be a semigroup satisfying the conditions of Deﬁnition 2.2. If a ∈ (G, ) and a a = a, then a = f where f is given by (iii).

Proof 1 By (iv),ifa ∈ G we can ﬁnd b ∈ G satisfying b a = f , so by (iii)

a = f a = (b a) a = b (a a) = b a = f,

by associativity, the hypothesis and (iv) again.

Theorem 2.5 A semigroup (G, ) satisfying Conditions (iii) and (iv) in Deﬁni- tion 2.2 forms a group as given by Deﬁnition 2.2.

Proof We need to show that f , and the inverses, apply both on the left and on the right, and are unique; that is, f as the neutral element, and h as the inverse of g. First, we show that if a ∈ (G, ) and b a = f , then a b = f (a left inverse is also a right inverse). We have

b (a b) = (b a) b = f b = b.

by (ii), (iv), and (iii). Hence, by (ii) again

(a b) (a b) = a (b (a b)) = a b.

By Lemma 2.4, this shows that a b = f ; the ﬁrst part follows. Secondly f is a right identity. We have, using the above subresult and (ii),

a f = a (b a) = (a b) a = f a = a

by (iii). Thirdly, we show that b is unique (that is, inverses are unique). For suppose c a = f , then we have by the above and (ii)

c = c f = c (a b) = (c a) b = f b = b,

1To emphasise their importance, and to aid clarity, all proofs are typeset indented. 2.1 Basic Deﬁnitions 15

by hypothesis and (iii) applied to b. Lastly, the neutral element. For if e also satisﬁes (iii), that is e a = a for a ∈ G, then substituting e for a we obtain e e = e, and so, by Lemma 2.4, e = f . This completes the proof.

From now on, we adopt the following conventions. We write ab for a b, e for the neutral element, and G for (G, ) when it is clear which operation is being used. Also, the inverse of g given by (iv) in Definition 2.2 will be written in the standard notation g−1 (and −g if we are using addition). We normally drop brackets and write xyz for either x(yz),or(xy)z. In some cases, we do not delete the brackets if this aids clarity. The next three results apply to all groups, and they will often be used in the sequel usually without being specifically identified. Note that no restrictions apply, a rare occurrence in the theory!

Theorem 2.6 (Cancellation) Suppose a,b,x,y ∈ G. If ax = bx, or if ya = yb, then a = b.

Proof From ax = bx we obtain, by Deﬁnition 2.2 and associativity, a = ae = a xx−1 = (ax)x−1 = (bx)x−1 = b xx−1 = b.

A similar argument applies in the second case.

Theorem 2.7 Suppose a and b are elements of a group G. (i) (ab)−1 = b−1a−1. (ii) (a−1)−1 = a. (iii) If G is ﬁnite, then a−1 equals some positive power of a. (iv) If a commutes with b, then a−1 also commutes with b.

Proof (i) As (b−1a−1)(ab) = b−1(a−1a)b = b−1b = e and, by Theorem 2.5, inverses are unique and two-sided, it follows that b−1a−1 is the inverse of ab. A similar argument applies for (ii). (iii) If G has n elements and a = e, then an integer m must exist satisfying 1

··· −1 = −1 ··· −1 Using induction we can extend (i) to prove that (a1 an) an a1 .

Theorem 2.8 If we treat the group G as a set (that is, we consider the underlying set of G) then, for all ﬁxed a ∈ G,

G ={ag : g ∈ G}={g−1 : g ∈ G}.

Proof Use Theorems 2.6 and 2.7; see Problem 2.1. 16 2 Elementary Group Properties

Most elementary exponent properties apply to groups. Note that as the group in question may not be Abelian some properties do not always hold. For example, (ab)2 may, or may not, equal a2b2. For all elements a in a group, if n ≥ 0 we write

+ a0 = e and an 1 = ana, that is an = aa ···a(ncopies of a).

Also, again if n ≥ 0 we write a−n in place of (a−1)n. By Theorem 2.7,thisalso equals (an)−1; reader, why?

Theorem 2.9 Suppose a is a group element, and r, s ∈ Z. (i) ar+s = ar as . (ii) (ar )s = ars.

Proof (i) By induction on s. Suppose s is non-negative. We have ar+0 = ar = ar e = ar a0 and, using the inductive hypothesis in the third equation,

ar+(s+1) = a(r+s)+1 = ar+sa = ar asa = ar as+1.

Now apply induction. Using this we have ar−sas = a(r−s)+s = ar , hence ar−s = ar (as)−1 = ar a−s and (i) follows for negative s. (ii) Assume ﬁrst that s is non-negative. As above we use induction on s, we have (ar )0 = e = a0r and

+ ar (s 1) = ar sar = arsar = ar(s+1),

and this case follows by the inductive hypothesis and (i). The reader should do the remaining case using a similar method to that given in the last part of the proof of (i).

We shall see below that an important invariant of a group is the number of elements in its underlying set, we deﬁne this as follows.

Deﬁnition 2.10 (i) Two groups G and H are called equal, G = H , if and only if their underlying sets are equal (page 277), and they have the same operation. (ii) The order of a group G is the number (or cardinality) of elements in the underlying set of G, this is denoted by o(G).

Some comments on cardinality are given in Appendix A. One or two authors reserve the word ‘order’ for groups and use the word ‘size’ for sets, we shall use ‘order’ for both. Note that o(G) can be finite or infinite, and this distinction is important; see page 5. If the order of G is finite, then the usual number-theoretic rules apply and, as we shall show later, they have a powerful controlling influence on the structure of G.Ifo(G) is infinite, then different considerations apply and care is needed when interpreting results. 2.2 Examples 17

Isomorphism—A Preliminary Note

Several groups can appear to be distinct but are, in fact, identical from the group- theoretical point of view. If we have two groups G1 and G2 with a bijection θ between their underlying sets which preserves or transforms all group-theoretic properties of G1 to G2, and vice versa, then we say they are isomorphic, and θ is an isomorphism between them, symbolically this is written G1 G2. The main property is (ab)θ = aθ · bθ, (2.1) for all a,b ∈ G1; see page 68. We shall give a formal deﬁnition of this concept at the beginning of Chapter 4, but it will be convenient to be able to use this notion from now on. As an illustration, we give two examples here. If G = H ,seeDef- inition 2.10(i), then the identity map (page 281) clearly acts as an isomorphism. Secondly, the real numbers with addition R, and the positive reals with multiplication R+, both form groups. They are isomorphic, and one isomorphism θ deﬁned by

xθ = 2x, for x ∈ R, : R → R+ demonstrates this fact. The map θ is a bijection (with inverse log2) and it transfers all group-theoretic properties of the ﬁrst group to the second, and vice versa via (2.1). For instance, the neutral element 0 of R is mapped to 20 = 1, the neutral element of R+, and if a,b ∈ R then (a + b)θ = 2a+b = 2a2b = aθbθ which veriﬁes (2.1) in this case.

Isomorphism Class

Consider the statement: “There are only two groups of order 6” (Problem 2.20). This is not correct as it stands because there are inﬁnitely many distinct groups of order 6, but many are isomorphic to one another. So to be more precise, we should say that “there are exactly two isomorphism classes of groups of order 6”. If we take the group with elements {0, 1, 2, 3, 4, 5} and operation addition modulo 6 (Z/6Z,the cyclic group of order 6), and D3 (page 3) as our ‘standard’ groups of order 6, then it is true that all groups of order 6 are isomorphic to one of these two groups. When discussing groups of a ﬁxed size we shall often use this short-hand.

2.2 Examples

Groups are found throughout mathematics, there is hardly a branch of the subject where they do not occur, they are also widely used in many branches of the physical sciences. We give here an extensive list of examples to illustrate the range and applicability of the group concept. No proofs will be given, in most cases it is not 18 2 Elementary Group Properties difﬁcult to check that the group axioms are satisﬁed. Note that the notation for individual groups given in this section will be used throughout the book, see pages 303 and 304.

Number Systems

Our first examples are the standard number systems. The integers Z, the rational numbers Q, the real numbers R, and the complex numbers C, with the operation of standard addition in each case, all form infinite Abelian groups. The non-zero rational numbers Q∗, non-zero real numbers R∗, and non-zero complex numbers C∗, with the operation of multiplication in each case, also form infinite Abelian groups distinct from the above. In general, for a ring or field F we let F ∗ denote the multiplicative group of the non-zero elements of F . Further, the positive rationals Q+ with multiplication form a group (which is a subgroup of Q∗; see Section 2.3), with a similar construction for R+. Note that neither the non-zero integers with multiplication nor the positive integers with multiplication form groups as inverses do not exist.

Modular Arithmetic

Our second collection of examples are ﬁnite groups from number theory. If m>0, the congruence a ≡ b(mod m) stands for: a and b have the same remainder after division by m (in symbols, m | b − a). This notation was ﬁrst introduced by C.F. Gauss in 1801 in his famous number theory text called ‘Disquisitiones arithmeticae’. Let Z/mZ denote the set {0, 1,...,m− 1}.Ifa,b ∈ Z/mZ, the operation +m is given by

a +m b = a + b, if a + b

a +m b = a + b − m, if a + b ≥ m,

(so a +m b ≡ a + b(mod m), this is called addition modulo m). The set Z/mZ with the operation +m is an Abelian group of order m, the notation Z/mZ which relates to cosets and factor groups will be explained in Chapter 4. This implies that at least one group of order m exists for each positive integer m; in some cases this is essentially the only group of this order (that is, up to isomorphism); for example, when m = 13 or 15, see Appendix D. If m is a prime number p, then (Z/pZ)∗ = (Z/pZ)\0 with multiplication modulo p, deﬁned similarly to addition modulo p, forms another ﬁnite Abelian group. Inverses exist by the Euclidean Algorithm (Theorem B.2 in Appendix B). Also note that the group T1 ={−1, 1} (page 42) is isomorphic both to the group Z/2Z, and to the group (Z/3Z)∗. 2.2 Examples 19

Product Groups

Given groups G1,...,Gn, we can form a new group by taking all (ordered) n-tuples of the form (g1,...,gn), where gi ∈ Gi for i = 1,...,n, as the new elements, and defining the new operation component-wise using the operations of each Gi in turn. In many cases, several different operations can be defined; see Chapter 7. For example, suppose n = 2 and G1 = G2 = T1. The elements of the product group are (1, 1), (1, −1), (−1, 1), (−1, −1), and the operation is given by (a, b)(c, d) = (ac,bd). This group is called the 4-group and is denoted by T2, it is a product of two copies of T1; in Chapter 7, we use the standard notation C2 × C2 for this group. Some authors use K (for Klein group) or V (for ‘Viergruppe’ the German word for ‘4-group’ or ‘fours group’) for this group. Note that the square of every element in T2 is the neutral element (1, 1), and it is an example of an Elementary Abelian Group as defined in Problem 4.18.

Matrix Groups

Matrix groups form one of the most important collections in the theory. Let F be a field (for instance, the rational numbers Q) and let m ≥ 1. The set of non-singular m × m matrices with entries from F and operation matrix multiplication forms a non-Abelian group (if m>1) called the m × m general linear group over F ,itis denoted by GLm(F ). The group axioms can be shown to hold using some elementary matrix algebra; the matrices are non-singular, and so inverses exist by definition. See Section 3.3 for further details. As we shall show later, subgroups (Section 2.3) of these matrix groups provide a further wide range of examples. For instance, (a) by considering only those matrices with determinant 1 in GLm(F ) we obtain the m × m special linear group denoted by SLm(F ), or (b) by considering those matrices that have zeros at all entries below the main diagonal we obtain the group of m × m upper triangular matrices denoted by UTm(F ); see Section 3.3. Also many examples can be obtained by choosing different fields F .SoifF is finite, these matrix groups provide a variety of examples of finite non-Abelian groups. For instance, SL2(F4) (which we usually write as SL2(4), the group of all 2 × 2 matrices A with det A = 1 and entries in the four element field F4—see page 254) is an important example of a simple group; as given by Definition 2.33. Many other simple groups can be defined using similar constructions; see Sections 12.2 and 12.3.

Symmetries of Geometric Objects

The symmetry properties of geometric objects provide a number of group examples. In Chapter 1 (page 3), we discussed the symmetries of an equilateral triangle, the 20 2 Elementary Group Properties group in question being called the dihedral group of the triangle denoted by D3.The elements of the group are the rotations and reflections that give the same geometric figure, and the operation is composition. A similar construction can be carried out for a regular polygon with n sides: the clockwise rotations about the centre are now by 2π/n, and ‘reflection’ or ‘turning over’ is as before. This group is denoted by Dn and again called dihedral. For example, D4 is the group of symmetries of the square, it has order 8. (Note that a few authors write D2n for Dn, see page 303.) Some other regular geometric objects have non-neutral (rotational) symmetry groups, for example, the tetrahedron (A4, see Problem 3.10), the octahedron (S4, see page 170) and the dodecahedron (A5, see Section 3.2 and Web Section 3.6). Also under this heading is the topic of sphere packing in various dimensions. Consider a large container filled with identical balls, some will touch adjacent balls and some will not. In dimension 2, where we have identical discs, a regular pattern forms and the set of disc centres gives a lattice (of equilateral triangles), and we can consider the symmetries of this lattice just as we have done for the triangle. In dimension 3, no such regular pattern forms where all adjacent balls touch. In this case, there are infinitely many ways to fit twelve balls around a central ball all touching it (there is always some room to spare), but thirteen never quite fit. In dimensions 8 and 24, ‘regular touching’ patterns do again form, the lattices given by the centres of the ‘spheres’ have some remarkable properties and give rise to some remarkable groups. For further details, see Conway and Sloane (1993). As a preliminary to this you should consider the following. The kissing number for these lattices is the maximum number of spheres that can fit around a central sphere S so that every sphere touches (kisses) S. In dimension 2, the kissing number is well- known to be six, and in dimension 3 it is, as noted above, twelve with some room to spare. But in dimension 8 it is 240, and in dimension 24 it is 196560 ! The first of these lattices has connections with the Mathieu group M24, and the second with the sporadic group called the ‘Monster’ or ‘Friendly Giant’, see Chapter 12,theATLAS (1985), and the reference quoted above.

Permutations

Permutations play a vital role in group theory, especially in the early development. If X is a set and SX denotes the collection of all permutations on X (that is, bijections of X to itself), then this collection forms a group under the operation of composition called the symmetric group on X.IfX is ﬁnite with n elements, we usually take X to be the set {1, 2,...,n} and write Sn for SX. See page 12 for the case n = 3. Note that Sn is non-Abelian if n>2, and has order n! (count all possible maps). As with many other groups, the symmetric groups have a number of important subgroups, that is, subsets that form groups; see Deﬁnition 2.11. For example, the alternating group An which is the group contained in Sn of all even permutations (a permutation is even if it can be expressed as an even number of interchanges of pairs of symbols; see Section 3.1) 2.2 Examples 21

Examples from Analysis

Some classes of functions form groups. For example, let Z denote the set of all continuous, strictly-increasing functions f which map [0, 1] onto [0, 1], and satisfy f(0) = 0 and f(1) = 1. This set Z forms a group if the operation is taken to be composition of functions (the identity function f0, where f0(x) = x for all x, acts as the neutral element, and inverses exist as the functions f are continuous and strictly monotonic). We can construct further groups (subgroups) inside this one, for instance, we could consider only those functions in Z which are differentiable. These are examples of ‘topological groups’, see page 6.

Free Groups and Presentations

This construction provides another way to introduce groups, it will be discussed in more detail in Section 3.4 and Web Section 4.7. Consider an alphabet of letters A ={a,a,b,b,...}. The letter a is going to act as the inverse of a, et cetera, see page 57.Aword c1c2 ···ck consists of a ﬁnite string of letters ci from the alphabet A, for example,

aabb a ,b,or ababa are words. The set of words with the operation of concatenation forms a semigroup; to obtain a group we proceed as follows. We define a reduced word as a word in which all pairs of consecutive letters aa,aa,bb,... do not occur or have been removed, for example, aabba reduces to a, whilst b and ababa are reduced. As a will act as the inverse of a, et cetera, each of these removals corresponds to the use of axiom (iv) in Definition 2.2. The operation of the group is as for the semigroup, that is concatenation—write one reduced word and then write the second reduced word immediately following the first, except that the resulting word must be reduced by removing all pairs aa,aa,bb,... if they are formed by the concatenation, or by previous removals. For instance,

the product of ac b and b ca c is c.

The empty word—that is the word with no symbols from A which is written as e where e/∈ A—acts as the neutral element of the group, and inverses are constructed as in the example above—for instance, the inverse of aabcbc is cbcbaa. The group is denoted by a,b,c,... (in this notation it is assumed that the letters e,a,b,...are also present), and the letters a,b,c,...are the generators. It is called free because there are no constraints on possible words other than those ensuring the group properties hold; note that all free groups are necessarily infinite. A free group with just one generator a, say, is called an infinite cyclic group, it is isomorphic to Z and so we denote it either by a or by Z. Non-free groups have more condi- 22 2 Elementary Group Properties tions called relations, or sometimes defining relations. For example, the finite cyclic group Cn of order n can be treated as the infinite cyclic group Z a with the extra relation an = e. In this group, each time an occurs it is replaced by the neutral element e in the same way that terms of the form aa or aa are replaced by e. In Section 3.4, we shall see that this method for constructing groups has a number of advantages, but also some disadvantages. For instance, in a few cases it may be difficult, or sometimes impossible, to determine the group order.

Elliptic Curves

The collection of solutions of some equations can be formed into groups. For example, consider the set of rational solutions of the equation

y2 = x3 + k where k ∈ Z\{0}. (2.2)

Suppose P1 = (x1,y1) and P2 = (x2,y2) lie on the curve. The straight line P1P2 passing through the points P1 and P2 meets the curve in exactly one further point P3 = (x3,y3), say, and the pair (x3,y3) forms a new solution of (2.2), and if x1,...,y2 ∈ Q then also x3,y3 ∈ Q.IfP1 = P2, then the line is the tangent to the curve at P1; and the whole procedure is called the chordÐtangent process. Points on the curve with rational coordinates form the elements of a group, and the operation is deﬁned using the chord–tangent process. It is closed because, given rational points P1 and P2 on C, a rational point P3 on C always exists, and we set P1 + P2 =−P3. The neutral element is the ‘point at inﬁnity’ on the curve in the y-direction. (To set this procedure up properly we use homogeneous coordinates (x : y : z), where

∗ (tx : ty : tz)= (x : y : z) for all t ∈ Q .

The usual notation for a point (x, y) is identified with (x : y : 1), and the point (x : y : 0) lies on the ‘line at infinity’. Equation (2.2) becomes y2z = x3 +kz3. The point (0 : 1 : 0), the neutral element of the group, clearly lies on the curve, and is the ‘point at infinity’ in the y-axis direction. We set P1 + P2 to equal ‘minus’ P3 to obtain a valid inverse operation, so using the standard two variable (affine) coordinates, the inverse of the point P1 = (x1,y1) is −P1 = (x1, −y1).) In this ‘projective geometry’ all vertical lines ‘meet’ at the point at infinity (0 : 1 : 0), and some results from algebraic geometry are needed to prove associativity. These groups can be finite or infinite, and they are Abelian because the line through the points P1 and P2 is clearly the same as the line through P2 and P1. See for example Rose (1999), Chapters 15 and 16, for further details. 2.2 Examples 23

Examples from Topology

The basic structure of a topological space is best described using groups. For example, the fundamental group of a space, which was ﬁrst deﬁned by H. Poincaré over a century ago, is constructed as follows:

Fix a point P in a path-wise connected topological space T , and consider the set of all continuous closed and directed loops from P to P . Call two loops ‘equivalent’ if one can be continuously deformed into the other (topologists call them ‘homo- topic’), the group operation is composition. The neutral element is the set of loops that can be continuously contracted to the point P , and the inverse of the loop L is L with its direction reversed. For instance, the fundamental group of the real plane R2 with the origin removed is the inﬁnite cyclic group (all loops through P that do not enclose the origin can be contracted to P , but, for example, a loop that passes around the origin clockwise four times, say, cannot be contracted and ‘equals’ four times a loop which passes around the origin clockwise only once).

Although the deﬁnition appears to depend on the point P ∈ T , it can be shown that fundamental groups for different points P are isomorphic; that is, there exists a fundamental group for the space T . Further details can be found in a standard book on general topology, for example, Kelley (1955) or Willard (1970). Algebraic topology is another area where groups—homology and cohomology groups-provide insights into the structure of topological spaces, see for example Benson (1991).

Examples from the Physical Sciences

Particle physicists make extensive use of group theory. Many of the essential properties of the basic constituents of matter are best described using the language and properties of groups. At least one elementary particle was discovered using the abstract theory. A collection of particles was associated with a particular class of groups, and it was realised that there was one more group (that is, 28) than known particles in this collection (at the time, 27); this led the experimental workers to look for the ‘missing’ particle (called Ω−), and it was duly found a few years later! An excellent ‘down-to-earth’ account of this topic is given in Close (2006), and Williams (1994) provides a good technical introduction. Some chemists use groups to describe the structure of molecules. A notable example was given in 1985 when a crystalline form of carbon, called Carbon60, was discovered by the chemists Kroto, Curl and Smalley;2 its structure is closely related to that of a dodecahedron, and so also to the alternating group A5; the subsection on symmetries above, Section 3.2, and Web Section 3.6 all give some details.

2They were awarded the Nobel prize in chemistry for this work. 24 2 Elementary Group Properties

2.3 Subgroups, Cosets and Lagrange’s Theorem

Most groups contain a number of smaller groups using the same operation, we shall consider these now.

Deﬁnition 2.11 A subgroup H of a group G is a non-empty subset of G which forms a group under the operation of G.

We write H ≤ G when H is a subgroup of G. For example, if G is the group Q, then Z is a subgroup, that is, Z ≤ Q. But note that Q+ is not a subgroup of Q even though the underlying set in the ﬁrst group is contained in the second; reader, why?

Deﬁnition 2.12 (i) A subgroup J of a group G is called proper if J = G,thisis denoted by J

Notes (a) The neutral subgroup e is sometimes called the identity, trivial, or unit, subgroup; see page 4. (b) Clearly G ≤ G; so all groups with more than one element have at least two subgroups; some have no more, see Theorem 2.34. (c) Maximal subgroups are not necessarily ‘large’. For an extreme example, consider the alternating group A13 which has order 3113510400, remarkably it possesses a maximal subgroup of order 78. Also arbitrarily large groups with maximal subgroups of order 2 exist—Problems 3.20 and Corollary 6.12. (d) There are connections between maximal subgroups and generators, see Prob- lem 2.13 and Section 10.2, and reasoning with maximal subgroups is used in several proofs, for example, in that for the Frattini Argument (Lemma 6.14).

The next result gives conditions for a group subset to be a subgroup.

Theorem 2.13 If H is a subset of G, then H ≤ G if, and only if, (a) H is non-empty, and (b) whenever a,b ∈ H , we also have a−1b ∈ H .

Proof Clearly (a) and (b) are valid if H ≤ G (Deﬁnitions 2.2 and 2.11). Conversely, suppose (a) and (b) hold for a subset H of G. By (a), there is at least one element a ∈ H , and so, by (b), e = a−1a ∈ H . Applying (b) again, we have, as a,e ∈ H , a−1 = a−1e ∈ H , and so H is closed under inverses. Thirdly, if a,b ∈ H , then a−1 ∈ H , and so together these give 2.3 Subgroups, Cosets and Lagrange’s Theorem 25

ab = (a−1)−1b ∈ H by Theorem 2.7(ii), that is, H is closed under the operation of G. Finally, we note that associativity holds in H because it holds in G; the result follows.

There is also a ‘right-hand version’ of this result where in (b) ‘a−1b ∈ H ’issubsti- tuted by ‘ab−1 ∈ H ’. In practice, it is often better to replace (b) by: (b1) if a,b ∈ H then ab ∈ H , and (b2) if a ∈ H then a−1 ∈ H .

For example, suppose G = GL2(Q) and H is the subset of these matrices with determinant 1. The identity matrix I2 belongs to H , and so H is not empty. Also if A,B ∈ H , then det A = det B = 1 and so, as det(AB) = det(A) det(B) = 1, we deduce AB ∈ H . Finally, if C ∈ H , then 1 = det C = det C−1, and so C−1 ∈ H ;this gives H ≤ G. We use the notation SL2(Q) for H , see Section 3.3. Some further examples are given in Problem 2.10. We consider now some set-theoretic operations on subgroups.

Corollary 2.14 (i) If H ≤ J and J ≤ G, then H ≤ G, that is the subgroup relation is transitive. (ii) If H,J ≤ G and H ⊆ J , then H ≤ J .

Proof Both of these results follow from Theorem 2.13, see Problem 2.5.

Intersections of subgroups are always subgroups (but unions are usually not subgroups because closure fails). See the note concerning subgroup lattices on page 32.

Theorem 2.15 Suppose I is a non-empty index set. If Hi ≤ G, for each i ∈ I , and = ≤ J i∈I Hi , then J G.

Proof As e ∈ Hi for all i ∈ I ,wehavee ∈ J ,soJ is not empty. Secondly, if a,b ∈ J , then a,b ∈ Hi for all i ∈ I , but each Hi ≤ G so, by Theorem 2.13, −1 −1 a b ∈ Hi , for all i ∈ I , which shows that a b ∈ J . Now use Theorem 2.13 again.

In Section 2.1 (page 13), we introduced the notion of a generating set for a group, this can be formally deﬁned by

Deﬁnition 2.16 A subset X of the underlying set of a group G is said to generate G if the intersection of all subgroups of G that contain X coincides with G,or to put this another way, the only subgroup of G that contains X is G itself. This intersection is denoted by X.

Theorem 2.17 Suppose X is a non-empty subset of the underlying set of the group G. The set X generates G if and only if the set of all products of powers (positive and negative) of elements of X equals G. 26 2 Elementary Group Properties

Proof By Theorem 2.15 and Definition 2.16, X≤G. Suppose X=G. Let Z denote the set of all powers of products of elements of X; Z ⊆ G). The set Z is non-empty as X is non-empty, and so Z ≤ G by Theorem 2.13 and the definition of Z.AlsoX ⊆ Z, and so Z is one of the subgroups used in the formation of the intersection X; hence X≤Z ≤ G. Therefore, as X generates G (by supposition), we have Z = G. For the converse suppose Z = G.NowforgivenH ,ifX ⊆ H and H ≤ G, then Z ≤ H , again by Theorem 2.13 and the definition of Z. This holds for all such H , and so it holds for X by Theorem 2.15; that is, Z ≤X. But by our supposition Z = G, and so X=G, and the result is proved.

We set X=e,ifX is empty. Now we consider group elements in more detail. For g ∈ G we write g for the setofpowersofg ∈ G, that is g={gt : t ∈ Z}; see Section 4.3.Wehave

Theorem 2.18 If g ∈ G then g≤G.

Proof The set g is clearly not empty, and if m, n ∈ Z, then gm,gn ∈g, and (gm)−1gn = gn−m ∈g. Result follows by Theorems 2.9 and 2.13.

We say that g is a generator of the subgroup g of G (page 13). This result ensures that almost all groups have at least some non-neutral proper subgroups, see Theorem 2.34 for the exceptions.

Examples (a) Let G = Z and g = 7, then 7 is the proper subgroup of Z consisting of the set of integers divisible by 7. (b) Secondly, let G = (Z/7Z)∗ and g = 3. In this case, the subgroup 3 is G itself because the powers of 3 modulo 7 generate the whole group; the reader should check this and also consider the case g = 2.

Theorem 2.18 and these examples suggest the following

Deﬁnition 2.19 Let g ∈ G. (i) The subgroup g given by Theorem 2.18 is called cyclic. (ii) The order of g, denoted by o(g), is deﬁned by o(g) = o(g); that is, o(g) equals the order of the cyclic subgroup generated by g in G. (iii) An element of order 2 is called an involution. (iv) The exponent, if it exists, of a group G is the least common multiple of the orders of all of the elements of G; that is, the least positive integer m with the property: gm = e for all g ∈ G.

Notes (a) All parts of this definition are relative to a fixed group G. (b) Orders can be finite or infinite, and if the orders of two elements are finite it does not follow that the order of their product is finite (Problem 2.7). 2.3 Subgroups, Cosets and Lagrange’s Theorem 27

(c) If G is finite, it has an exponent which is not greater than o(G). The group Q is an example of an infinite group with no exponent. In 1902, W. Burnside (1852– 1927) conjectured that a group G with finite generating set and finite exponent must be finite, and this is true if G is Abelian. But it can be false if G is not Abelian as was shown by Adian and Novikov in 1968 for a group with an exponent larger than 665; see Vaughan-Lee (1993). (d) Elements of order 2 are called involutions to signal the fact that they play a unique role in the theory, particularly to CFSG. (It is the subgroups called centralisers of these involutions, see Section 5.2, that play this vital role.) Also, apart from the neutral element e they are the only group elements which equal their own inverses. Further properties are given in Problems 2.28, 3.1(iv) and 3.20, and in the note about Coxeter Groups on page 64.

We illustrate these concepts with the following result.

Corollary 2.20 If a group G has exponent 2, then it is Abelian.

Proof Suppose a,b ∈ G, then ab ∈ G and e = (ab)2 = abab. Multiplying on the left by a and on the right by b, we obtain

ab = aeb = a(abab)b = a2bab2 = ba

as both a and b have order 2. This holds for all a,b ∈ G.

Given a prime p, an Abelian group all of whose non-neutral elements have order p is called an Elementary Abelian p-group. We shall see later (Problem 4.18) that these groups can be treated as vector spaces deﬁned over a p-element ﬁeld. The corollary above shows that all groups of exponent 2 are of this type, this is not true for primes p>2; an example is given in Problem 6.5.

Cosets and Lagrange’s Theorem

For our next results, we require some new notation. If X and Y are non-empty subsets of a group G, then we write XY for the subset

XY ={xy : x ∈ X and y ∈ Y }⊆G.

If X is the singleton set {x}, then we write xY for {x}Y (and Yx for Y {x}). Note that if X, Y, Z ⊆ G then, by associativity, (i) X(YZ) = (XY )Z, and (ii) XY = YX,ifG is Abelian.

Deﬁnition 2.21 For H ≤ G and g ∈ G,thesetgH ={gh : h ∈ H } is called a left coset of H in G, and the set Hg is called a right coset of H in G. 28 2 Elementary Group Properties

Cosets play an important role in the theory, here they lead to our ﬁrst major result—Lagrange’s Theorem, and in Chapter 4 they form part of the important ideas associated with factor groups. One of the origins of this work is Gauss’s development of modular arithmetic undertaken two centuries ago (page 18): if G = Z and H = nZ, then the cosets are

kH ={k + nz : z ∈ Z} for k = 0,...,n− 1; the coset kH equals the set of integers congruent to k modulo n. When referring to the set T of cosets of H in G, we often write T ={gH : g ∈ G}. Here we are using the convention that in an un-ordered set duplication is ignored, for instance, the set {...,a,a,...,a,...} is the same as {...,a,...}.Ifwe did not use this convention in the coset case, we would need to specify a unique g in each coset gH, and this would cause problems. We begin with some basic lemmas. The ﬁrst will be used often in the following pages, it characterises the coset representatives.

Lemma 2.22 If H ≤ G and a,b ∈ G, then

aH = bH if and only if a ∈ bH if and only if b−1a ∈ H.

There is an exactly similar result for right cosets; the reader should write it out and redo the following proof in this second case.

Proof Suppose ﬁrstly aH = bH.AsH is a subgroup, e ∈ H and so a = ae ∈ aH = bH. Secondly, suppose a ∈ bH, then there exists h ∈ H satisfying a = bh, and so b−1a = h ∈ H . Lastly, suppose b−1a ∈ H . As above this gives a = bh for some h ∈ H , and hence

ah1 = bhh1 ∈ bH for all h1 ∈ H ;

that is, aH ⊆ bH. For the converse inclusion, as H ≤ G, we have by Theo- rem 2.13, a−1b = (b−1a)−1 ∈ H , and so we can repeat the previous argument with a and b interchanged, the equation aH = bH follows.

To derive Lagrange’s Theorem, we require the following three lemmas, the ﬁrst shows that cosets are either disjoint or identical.

Lemma 2.23 If H ≤ G, then the underlying set of G can be expressed as a disjoint union of the collection of all left cosets of H in G. There is an exactly similar result for right cosets.

Proof Clearly, each element g ∈ G belongs to a left coset because g ∈ gH. Suppose further g ∈ aH and g ∈ bH, then by Lemma 2.22, aH = gH = bH, and the lemma follows. The right coset version follows similarly. 2.3 Subgroups, Cosets and Lagrange’s Theorem 29

Lemma 2.24 If H ≤ G and g ∈ G, then o(H) = o(gH) = o(Hg).

Proof We give the proof for left cosets, the right coset result is proved similarly. To establish this lemma, we construct a bijection between the sets involved. Let φ be the map from H to gH deﬁned by

hφ = gh;

see note on ‘left or right’ on page 68.Ifhφ = hφ then gh = gh, and by cancellation (Theorem 2.6) this gives h = h, and so φ is injective, hence it is bijective because it is clearly surjective.

Lemma 2.25 If H ≤ G, then the number (cardinality) of left cosets equals the number of right cosets.

Proof As in the previous proof, we construct a bijection between the sets. Let θ be the map from the set of left cosets to the set of right cosets given by

(gH )θ = Hg−1.

= −1 = −1 This is well-deﬁned; for if gH g1H , then Hg Hg1 (use left and right versions of Lemma 2.22 and closure under inverses). It is also clearly surjective because each element of G is the inverse of some element in G (Theorem 2.8). To prove injectivity, suppose = −1 = −1 (gH )θ (g1H)θ, that is Hg Hg1 . −1 ∈ −1 Using the right-hand version of Lemma 2.22, this gives g1 Hg , and −1 = −1 ∈ = −1 ∈ hence g1 hg for some h H . Therefore, g1 gh gH (as H is a subgroup), and so g1H = gH by Lemma 2.22 again.

Having established these lemmas, we can now derive Lagrange’s Theorem. Much of the early work in algebra was concerned with properties of polynomials defined over the rational numbers. J.-L. Lagrange (1736–1813), an Italian mathematician working in France, studied these polynomials as well as in many other topics in mathematics. He postulated a result similar to that given in the last part of Prob- lem 5.1 which relies on what we now call ‘Lagrange’s Theorem’; and it is for this reason that the following result is so named. In fact, Galois gave the first proof of the theorem for permutation groups in 1832, and it was probably C. Jordan (1838–1932) who gave the first proof for general groups some thirty years later. We begin by making the following

Deﬁnition 2.26 Let H ≤ G. The number (cardinality) of left cosets of H in G is called the index of H in G, it is denoted by [G : H].

By Lemma 2.25, this equals the number of right cosets of H in G. 30 2 Elementary Group Properties

Theorem 2.27 (Lagrange’s Theorem) If H ≤ G then o(G) = o(H)[G : H ].

Proof By Lemma 2.23, the underlying set of G is a disjoint union of [G : H ] cosets, and by Lemma 2.24, each of these cosets has the same cardinality (number of elements), that is o(H), the theorem follows.

This result is particularly useful in the finite case where it shows that H can only be a subgroup of G if o(H) | o(G); that is, the prime factorisation of the order of a group G is an important invariant of G. For instance, a group of order 30 cannot have subgroups of order 4, 7, 8, 9, 11,...,29. Also, it cannot have elements of order 4, 7,..., see Definition 2.19. In the infinite case, the theorem shows that either the order of the subgroup, or the index (or both), must be infinite. Note that there exist infinite groups all of whose proper subgroups are finite, see Problem 6.7.

2.4 Normal Subgroups

The last topic in this chapter concerns a special type of subgroup in which left and right cosets are equal, they play a vital role in the theory. These subgroups were ﬁrst deﬁned by Galois in the 1820s when he was working on the solution of polynomial equations by radicals; see the Introduction to Chapter 11.

Definition 2.28 (i) Let K ≤ G. The subgroup K is called normal in G if, and only if, gK = Kg for all g ∈ G. This is denoted by K G. (ii) If g,h ∈ G, h−1gh is called the conjugate of g by h in the group G. (iii) For a fixed element g ∈ G,theset{h−1gh : h ∈ G}, that is, the set of conjugates of g in G, is called the conjugacy class of g in G; see also Definition 5.17.

Notes (a) The subgroups e and G are normal in G for all groups G. (b) If G is Abelian, all subgroups are normal, all conjugates of g equal g, and so the conjugacy class of g in G is {g}. (c) We reserve the symbol ‘K’, possibly with primes or subscripts, to denote a normal subgroup, but other symbols will occasionally be used where necessary. In Chapter 4, we discuss the connection between normal subgroups and kernels of homomorphisms—K for kernels and so also for normal subgroups. (d) Conditions stronger than normality are useful at times. The first is characteristic, for a definition and basic properties see Problem 4.22; the main point is that characteristic is a transitive property whereas normality is not. Some authors use an even stronger property called fully invariant which is defined similarly to characteristic except that in the definition on page 89 the word ‘automorphism’ is replaced by ‘endomorphism’, see Definition 4.2. 2.4 Normal Subgroups 31

The following theorem gives two conditions for normality, see note below the statement of Lemma 4.6.

Theorem 2.29 (i) If K ≤ G, then the following conditions are equivalent: (ia) K G; (ib) for all g ∈ G, g−1Kg ⊆ K; (ic) for all g ∈ G and all k ∈ K, g−1kg ∈ K. (ii) Suppose K G. If k ∈ K, then all conjugates of k in G belong to K, and K is the union of a collection of the conjugacy classes of G.

Proof Note first that both parts of (ii) follow immediately from (i). Suppose (ia) holds, so if g ∈ G, gK = Kg by definition. Hence, for all k ∈ K, we can find k ∈ K to satisfy

− gk = kg, that is g 1kg = k ∈ K,

which gives (ib). Secondly, note that (ic) follows immediately from (ib) (as g−1kg ∈ g−1Kg). Finally, suppose (ic) holds. So if g ∈ G and k ∈ K, we can ﬁnd k ∈ K to satisfy

− g 1kg = k , which gives kg = gk and so Kg ⊆ gK,

as this argument holds for all k ∈ K. For the converse, we have gkg−1 = (g−1)−1kg−1 ∈ K, and so we can ﬁnd k ∈ K to satisfy gkg−1 = k or gk = kg. This gives the reverse inclusion and (ia) follows.

Notes To prove that K G it is necessary to prove both K ≤ G and K is normal in G. Secondly, normality is not transitive (cf. Corollary 2.14); that is, if K G and H K, it does not follow that H G; see Problem 2.19(iii) for an example. On the other hand, if K G and K ≤ H ≤ G, then K H ; see Problem 2.14. A stronger property called characteristic which is transitive was mentioned in (d) opposite.

Our ﬁrst application of the normal subgroup concept answers the question: When is the product HJ of two subgroups H and J itself a subgroup? Note that in general HJ is not a subgroup because it is not closed under the group operation. We use the notation H ∨ J (or sometimes H,J)—the join of H and J —for the group generated by the elements of both H and J (Deﬁnition 2.16). Clearly HJ ⊆ H ∨ J , we have

Theorem 2.30 Suppose H,J ≤ G. (i) If either H or J is a normal subgroup of G, then HJ ≤ G and H ∨ J = HJ = JH. (ii) If both H G and J G, then HJ G. 32 2 Elementary Group Properties

Proof (i) Suppose hi ∈ H,ji ∈ J , i = 1, 2, and H G (the proof is similar −1 −1 = ∗ ∗ ∈ −1 ∈ ∈ if J G). Then j1 (h1 h2)j1 h for some h H (as h1 h2 H,j1 J and H G). Hence −1 = −1 −1 −1 = ∗ −1 ∈ (h1j1) (h2j2) j1 h1 h2j1j1 j2 h j1 j2 HJ, and, as HJ is clearly not empty, HJ ≤ G follows by Theorem 2.13.Asim- ilar argument shows that a product of terms each of the form hj ,forh ∈ H and j ∈ J , is itself of this form; that is, HJ = H ∨ J . The last equation in (i) follows because H ∨ J = J ∨ H , or we can show directly as above that HJ ⊆ JH and JH ⊆ HJ. (ii) By (i), we only need to check normality. If g ∈ G, h ∈ H and j ∈ J , we have g−1hj g = g−1hgg−1jg ∈ HJ by hypothesis, the result follows.

Subgroup Lattices

Using Corollary 2.14 and Theorems 2.15 and 2.30, the collection of subgroups of a group forms a (complete) lattice L, that is, a non-empty partially ordered set (Def- inition A.5 in Appendix A) in which every subset has a greatest lower bound and a least upper bound in L. Note that both the intersection and the join of two subgroups of a group G are themselves subgroups of G. Some examples are given in Chapter 8. The structure of this lattice can have an important bearing on the group in question. The ﬁrst major result (Ore 1938) states that the lattice of a ﬁnite group G is distributive (that is, H ∨ (J ∩ K) = (H ∨ J)∩ (H ∨ K), et cetera.) if and only if G is cyclic. It should be noted that non-isomorphic groups can have identical subgroup lattices; see Problem 6.4. For a detailed account of this aspect of the theory, the reader should consult Schmidt (1994). The centre of a group is an important example of a normal subgroup, it is given by

Lemma 2.31 In a group G, the set

J ={a ∈ G : ag = ga for all g ∈ G} forms a normal Abelian subgroup of G.

Proof Suppose a ∈ J . For all g ∈ G,wehaveeg = ge, ag = ga implies a−1g−1 = g−1a−1 by Theorem 2.7, and if ag = ga and bg = gb then abg = agb = gab; and so J ≤ G by Theorem 2.13.AlsoJ is Abelian by deﬁnition. Lastly, note that ag = ga implies g−1ag = a, and so J G by Theorem 2.29.

Deﬁnition 2.32 The subgroup J of G giveninLemma2.31 is called the centre of G, it is denoted by Z(G). 2.4 Normal Subgroups 33

Notes The notation Z(G) is used because German authors call this subgroup the Zentrum. The centre of a group G gives some important information about G. Clearly, Z(G) = G if and only if G is Abelian. On the other hand, some groups are centreless, that is, Z(G) =e; examples are D3 and S4, see Problem 2.26. A centreless group can in some ways be treated as the opposite of an Abelian group.

We end this chapter by introducing simple groups. We shall show later they can be treated as the basic ‘building blocks’ for the construction of all ﬁnite and some inﬁnite groups; see Chapter 9.

Deﬁnition 2.33 A group G is called simple if it contains no proper non-neutral normal subgroup.

The term ‘simple’ is perhaps not well-chosen because some simple groups are very complicated! But as noted above they can be used as the basic constituents of all groups; of course, they are all centreless. A full list of ﬁnite simple groups is now known; see the ATLAS (1985) and Chapter 12. Many simple groups are given by Lagrange’s Theorem for we have

Theorem 2.34 If o(G) is a prime number, then the group G is simple and cyclic.

For the converse see Theorem 9.6.

Proof By Lagrange’s Theorem (Theorem 2.27), the order of a subgroup of G divides o(G). But in this case, the only positive divisors of the integer o(G) are 1 and p, hence G has no proper non-neutral subgroups at all, and so clearly no proper non-neutral normal subgroups. Also by Theorem 2.18 and Deﬁnition 2.19, every element has order 1 or p. There is only one element, e, of order 1 (Lemma 2.4). Hence G has p − 1 elements of order p;leta be one of them. As a has p distinct powers (including p0 = e), it follows that all elements of G equal powers of a, and so G is cyclic.

In fact, ‘most’ simple groups (counted by the size of their orders) are of this type, that is Abelian (and cyclic). For example, there are 173 (isomorphism classes of) simple groups with order less than 1000 but only five are non-Abelian. The construction of non-Abelian simple groups is a much more difficult task, in the next chapter we introduce the first groups of this type—alternating groups, and more will be discussed in Chapter 12. These include a number of infinite classes of matrix groups, especially the linear groups Ln(q) and the unitary groups Un(q), and also 26 (!) so called sporadic groups. These groups range in size from 7920 84 (Mathieu group M11) to about 10 (Friendly giant M) and they have a wide variety of constructions. The existence of these non-Abelian simple groups is surely one of the most interesting and challenging aspects of the theory. 34 2 Elementary Group Properties

2.5 Problems

A number of the problems given below have important applications in the sequel. For an explanation of the symbols and , see page 9.

Problem 2.1 (i) Write out a proof of Theorem 2.8. (ii) Using induction on n, prove the generalised associativity law for groups: If g1,...,gn ∈ G, then all expressions formed by inserting or deleting brackets (in corresponding pairs) in the term g1 ···gn are equal.

Problem 2.2 Show that the following sets with operations form groups, and indicate which are Abelian. (i) Z/7Z with addition modulo 7. (ii) (Z/7Z)∗ with multiplication modulo 7. (iii) The set Q with the operation ∗ where, for a,b ∈ Q,wehavea ∗ b = a + b + 3. (iv) GL2(Q) with matrix multiplication, see also Problem 2.10 below. (v) The set of powers of products Q (that is, the group generated by) of the com- A = 01 B = 0 i plex matrices −10 and i 0 with the operation of matrix multiplication. What is the order of this group? (vi) Let R = R ∪{∞} where the symbol ∞ satisﬁes the usual naive rules: 1/0 =∞,1/∞=0, ∞/∞=1 and 1 −∞=∞=∞−1. Deﬁne six functions mapping R onto itself by:

1 f (x) = x, f (x) = ,f(x) = 1 − x, 1 2 x 3 1 x x − 1 f (x) = ,f(x) = ,f(x) = . 4 1 − x 5 x − 1 6 x Show that this set forms a ﬁnite group under the operation of composition. (vii) Let R denote the real plane R2,letd denote the standard distance function (metric) on R, and let denote the set of bijective maps of R to itself which preserve distance—if x,y ∈ R and θ ∈ , then d(x,y) = d(θ(x), θ(y)). A function of this type is called an isometry; rotation by π/3 about the origin is an example. Show that with the operation of composition forms a group.

The reader needs to be convinced that all the sets with operations described in Section 2.2 are, in fact, groups.

Problem 2.3 Why are the following sets with operations not groups? (i) The integers Z with subtraction. (ii) The set of odd integers with addition. ar ∪ c 0 ∈ R = = (iii) The set 0 b sd where a,b,...,s and ab 1 cd, with matrix multiplication. (iv) The rational numbers Q with multiplication. 2.5 Problems 35

Problem 2.4 (i) Let S be a semigroup with cancellation, so it has closure under its operation which is associative, and for all a,b ∈ S, we can ﬁnd x,y ∈ S to solve the equations ax = b and ya = b. Show that S forms a group. (ii) If T is a semigroup, and for all a ∈ T there is a unique a∗ ∈ T satisfying aa∗a = a, prove that T is a group.

Problem 2.5 If H,J ≤ G and in (iii) p is a prime, show that (i) If H is a subset of J , then H ≤ J . (ii) H ∩ J = H if, and only if, H ≤ J . (iii) If o(H) = o(J) = p, then either H = J or H ∩ J =e.

Problem 2.6 Prove that if G is a group and S ≤ G, then SS = S. Conversely, if T is a non-empty ﬁnite subset of G and TT = T , prove that T ≤ G.IsthistrueifT is inﬁnite?

Problem 2.7 (Order Function) Let g,h ∈ G. Prove the following properties of the order function. (i) o(gh) = o(hg). (ii) If o(g) = n and m ∈ Z, then o(gm) = n/(m, n); see page 284. (iii) If o(g) = m and (m, n) = 1, there exists h ∈ G satisfying hn = g. (iv) If o(g) = m, o(h) = n, and g and h commute, then o(gh) = LCM(m, n);see part (vii). (v) If G is finite and g ∈ G, then go(G) = e, and o(g) | ex(G) where ex denotes the exponent of G, see Definition 2.19. (vi) Suppose g ∈ G and o(g) = mn where (m, n) = 1. Show how to find unique a,b ∈ G to satisfy ab = g = ba, o(a) = m and o(b) = n. (vii) In (iv), if we drop the commutativity condition show that o(gh) can be infinite. (Hint. Try G = GL2(Q).)

Problem 2.8 (i) Suppose G is a ﬁnite group and o(G) is even. Is the number of elements of order 2 in G odd—does a group of even order always contain an involution? See also Cauchy’s Theorem (Theorem 6.2). (ii) Using the group (Z/pZ)∗ where p is prime, see the deﬁnition on page 18, give a proof of Fermat’s Theorem: − ap 1 ≡ 1 (mod p) if (a, p) = 1.

(iii) Using the same group as in (ii), show that (p − 1)!≡−1 (mod p),are- sult sometimes (wrongly) known as Wilson’s Theorem. You are given: If p>2, then (Z/pZ)∗ contains exactly two elements of order at most 2 (Theorem B.13 in Appendix B). 36 2 Elementary Group Properties

Problem 2.9 (Multiplication Tables) Given a group G of order n with elements g1,...,gn where g1 = e, we can form a square array or table, with n rows and n columns, whose (i, j)th entry is the product gigj . Show that each row and each column of this table is a permutation of the elements g1,...,gn. What can you say about the first row and first column? Is the converse true? That is, if we have a square array of elements such that each row and each column is a permutation of some fixed set, and the first row and column have the property mentioned above, does the corresponding array always form the multiplication table of a group?

Problem 2.10 Show that the following subsets are subgroups of the corresponding groups, and determine whether they are normal.

(i) The set {1, −1} in R∗. (ii) The set of permutations on Y ={1,...,6} which leave 3 ﬁxed in S6,thesetof all permutations on Y ; see Section 3.2. (iii) The subsets of GL2(Q) of matrices (a) which have determinant 1, and (b) which are upper triangular (that is, the bottom left-hand entry is zero); see page 19 and Section 3.3. (iv) The set of complex numbers with absolute value 1 in C∗. (v) The set of differentiable functions in the group Z described in the subsection on groups in analysis on page 21.

Problem 2.11 (i) Show that a ﬁnite subgroup of the multiplicative group of the complex numbers C∗ is cyclic. (Hint. Consider roots of unity.) (ii) Find as many subgroups as you can of the additive group of the rational numbers Q;seeWeb Section 7.5.

Problem 2.12 List the left and right cosets of the subgroups given in Problem 2.10; note that the last part is not easy!

Problem 2.13 (i) Can a subset of a group G be the left coset of two distinct subgroups of G? (ii) If G is ﬁnite and has a unique maximal subgroup H , show that it is cyclic. (Hint. Consider an element in G\H .)

Problem 2.14 (Normality Properties) Prove the following statements—all widely used in the sequel.

(i) If K G and K ≤ H ≤ G, then K H . (ii) A subgroup of the centre of a groupG is normal in G. = n (iii) If Ki G for i 1, 2,...,n, then i=1 Ki G. (iv) If H,J,K ≤ G and K J , then K ∩ H J ∩ H . 2.5 Problems 37

Problem 2.15 (i) Show that if [G : H] is ﬁnite and H ≤ J ≤ G, then

[G : H]=[G : J ][J : H ].

(ii) Prove that if H,J ≤ G with [G : H]=m and [G : J ]=n, then [G : H ∩ J ]≥ LCM(m, n), and equality occurs if, and only if, m and n are coprime.

Problem 2.16 (Derived Subgroup) If g,h ∈ G we set [g,h]=g−1h−1gh,itis called the commutator of g and h. Also the subgroup of G generated by the set of all products of powers of the commutators of G is called the derived (or commutator) subgroup of G, and it is denoted by G. In some cases, the set of commutators of a group does, in fact, form a subgroup of the group, but not always; for an example, seeRotman(1994), page 34. More generally, if H,J ≤ G,welet[H,J] denote the subgroup generated by all commutators of the form [h, j] where h ∈ H and j ∈ J ; so, for example, [G, G]=G. See also Problem 4.6(ii), and Section 11.1, especially page 234. (i) Show that G G. (ii) Find G when G is (a) Z,(b)D3, and (c) Q, see Problem 2.2(v). (iii) Prove that if J ≤ G and J ⊇ G, then J G—an important fact with many applications. (iv) Show that if K G and K ∩ G =e, then K ≤ Z(G), and so in particular K is Abelian. (v) Finally, prove that if K G and J =[K,G], then J ≤ K and J G.

Problem 2.17 (Commutator Identities) Prove the following identities where [a,b,c]=[[a,b],c] for a,b and c in the same group G. Identity (iv) is called the HallÐWitt Identity. (i) [b,a]=[a,b]−1, (ii) If a,b ∈ G, and both a and b commute with [a,b], show that

[ar ,bs]=[a,b]rs for r, s ∈ Z, − (ab)t = at bt [b,a]t(t 1)/2 if t ≥ 0.

(Use induction on t, (i), and the given relationship between G and Z(G).) (iii) [ab,c]=(b−1[a,c]b)[b,c] and [a,bc]=[a,c](c−1[a,b]c), (iv) b−1[a,b−1,c]bc−1[b,c−1,a]ca−1[c,a−1,b]a = e. (v) If a1,...,am,b1,...,bn ∈ G and H =a1,...,bn, then we can express [a1 ...am,b1 ...bn] as a product of conjugates of [ai,bj ] by some cij ∈ H . (vi) If H,J ≤ G where G =H,J, then [H,J] G.

Problem 2.18 Suppose A,B,C ≤ G and A ≤ B. Show that (i) B ∩ (AC) = A(B ∩ C), (ii) if G = AC then B = A(B ∩ C), (iii) if AC = BC and A ∩ C = B ∩ C, then A = B. 38 2 Elementary Group Properties

(Note that AC and/or BC may not be subgroups of G, also (i) and (ii) are sometimes known as Dedekind’s Modular Laws.) (iv) Now suppose A,B,C,D ≤ G where also AB, CD ≤ G. Show that if A ≤ D and C ≤ B then AB ∩ CD = AC(B ∩ D).

Problem 2.19 Let H,J ≤ G. Prove the following results. (i) If [G : H ]=2, then (a) H G, and (b) a2 ∈ H for all a ∈ G—facts we use many times. (ii) If G is ﬁnite and o(H ) > o(G)/2, then H = G—no ﬁnite group can have a proper subgroup of order larger than half the group order. Further, if G is also simple and J ≤ G, then o(J) ≤ o(G)/3. For large simple groups, the denominator 3 can be replaced by a much bigger integer; see example below Theorem 5.15, page 101. (iii) Show that normality is not transitive (that is if H J and J G, it does not follow that H G); one example occurs in D4 using (i). (iv) If H and J are proper subgroups of G, prove that there exists g ∈ G which does not belong to either H or J . (v) Show that HJ ≤ G,ifHJ = JH; cf. Theorem 2.30(i).

Problem 2.20 Using Corollary 2.20, Lagrange’s Theorem (Theorem 2.27) and Problem 2.5, show that up to isomorphism there are only two groups of order 4, and only two groups of order 6—that is, there are exactly two isomorphism classes of groups of order 4, and also exactly two of order 6. (Hint. For order 6, show that the group always contains an element of order 3.) See Problem 4.2(i), different methods to prove these facts are given in Chapters 5 and 6.

Problem 2.21 Let G be a group. If Hi ≤ G and [G : Hi] is ﬁnite for i = 1,...,n, show that n n G : Hi ≤ [G : Hi]. i=1 i=1 (Hint. Derive the result for n = 2ﬁrst.)

Problem 2.22 (Poincaré’s Theorem) Prove that the intersection of a finite number of subgroups of G, each with finite index, is itself a subgroup of G with finite index.

Problem 2.23 If H ≤ G and g ∈ G, then g−1Hg is called a conjugate subgroup of H (Deﬁnition 2.28). Prove the following statements: (i) g−1Hg ≤ G, (ii) o(g−1Hg)= o(H), (iii) g−1Hg ={j ∈ G : gjg−1 ∈ H}. 2.5 Problems 39

Problem 2.24 (Core of a Subgroup) If H ≤ G,thecore of H in G, core(H ),is deﬁned by − core(H ) = g 1Hg, g∈G see Section 5.2. Show that (i) core(H ) G, (ii) core(H ) is the join of all normal subgroups of G which are contained in H , (iii) core(H ) is the unique largest normal subgroup of G contained in H .

Problem 2.25 (Normal Closure of a Subgroup) If H ≤ G, then the normal closure H ∗ of H is deﬁned as the intersection of all normal subgroups of G which contain H . Show that (i) H ∗ G, (ii) H ∗ =g−1Hg : g ∈ G, (iii) H ∗ is the smallest normal subgroup of G containing H .

Problem 2.26 Find the centres of the following groups. (i) Integers Z, (ii) Dihedral group D4, (iii) Dihedral group D5, (iv) 2 × 2 General linear group GL2(Q), and (v) Permutation group S3.

Problem 2.27 Prove that if H,J ≤ G, then

o(H J )o(H ∩ J)= o(H)o(J).

One method is as follows. Deﬁne a map θ : H × J → HJ by (h, j)θ = hj . Show that if g = hj where h ∈ H and j ∈ J , then gθ−1 = ha, a−1j : a ∈ H ∩ J , by proving inclusion both ways round. Further, show that if (ha, a−1j)= (hb, b−1j) then a = b, and so o(gθ−1) = o(H ∩ J). Lastly, count ordered pairs using the property o(H × J)= o(H)o(J). Note that (a) HJ need not be a subgroup of G, and (b) a second proof of this result is given in Theorem 5.8.

Problem 2.28 Suppose G is a ﬁnite simple group of even order. Using Prob- lem 2.8, show that G is generated by its involutions. (Hint. Note that an involution is self-inverse.) By the Feit–Thompson Theorem (Chapters 11 and 12), this shows that all ﬁnite non-Abelian simple groups are generated by a set of their involutions. 40 2 Elementary Group Properties

Problem 2.29 (Double Cosets) Suppose H,J ≤ G and a ∈ G.ThesetHaJ = {haj : h ∈ H,j ∈ J } is called the double coset of a with respect to H and J . Show that (i) Each element of G belongs to exactly one double coset. (ii) G is the disjoint union of its double cosets. (iii) Each double coset (with respect to H and J ) is a union of right cosets of H , and a union of left cosets of J . (iv) The number of right cosets of H in the double coset HaJ is [J : J ∩ a−1Ha]. Hence

[G : H]= [J : c−1Hc] c∈C provided this sum is ﬁnite, where C is a set of double coset representatives for H and J . (v) Using the notation set up in Section 3.1,ifG = S4, a = (1, 2), H =(1, 2, 3), J1 =(1, 2, 3, 4) and J2 =(1, 4)(2, 3), write out the double cosets HaJi for i = 1, 2.

≤ ≤···≤ Problem 2.30 (i) Suppose J 1 J2 G, that is we have an inﬁnite sequence = ∞ ≤ of subgroups of G.LetJ i=1 Ji . Show that J G. Note that in general a union of subgroups is not itself a subgroup. (ii) In (i), if Ji is simple for inﬁnitely many i, show that J is also simple.

Problem 2.31 (Project) Whilst reading this book, list all those theorems which apply without restriction or caveat, for example, one of the ﬁrst is Cancellation (The- orem 2.6). Chapter 3 Group Construction and Representation

Bertrand Russell defined the integer ‘3’ as that property common to all sets having three elements, with similar definitions for other positive integers. The abstract entity ‘3’ has many representations in the myriad of sets with three elements.1 So it also is with groups. For example, the alternating group A5 to be introduced in Sec- tion 3.2 is defined as the group of even permutations on a five-element set—the first representation of this group we give is expressed in terms of permutations. But it has several other representations (as a matrix group or a symmetry group, et cetera, see Web Section 3.6). The point being that when we discuss an individual group, we almost always discuss a particular representation of the group, as a matrix, or permutation, or other type of group, and the corresponding ‘abstract’ group is that entity common to all of these representations—this is an important point to bear in mind when discussing individual groups. Also these ideas have led to a branch of the subject called “group representation theory” which we shall introduce in Web Chapter 13. There is one type of representation which comes close to the ‘abstract’ group, that is a ‘group presentation’ which was introduced briefly on page 21. The group is defined on an ‘alphabet’, the elements are the ‘words’ in this alphabet, and the operation is defined using concatenation. For instance, one presentation of the dihedral group D3 described on page 3 is a,b a3 = b2 = (ab)2 = e .

In this representation, the elements of D3 are given by the products of powers of a and b subject to the conditions given above. Assuming that we can apply the usual elementary group rules, it is easy to show that this system only has six members: e,a,a2,b,ab, and a2b, and that they ‘correspond’ to the six symmetries of the triangle described on page 3. We shall consider this topic in more detail in Section 3.4.

1Nowadays a more inductive deﬁnition is given using the integer zero and the successor function.

H.E. Rose, A Course on Finite Groups, 41 Universitext, DOI 10.1007/978-1-84882-889-6_3, © Springer-Verlag London Limited 2009 42 3 Group Construction and Representation

In this chapter, we discuss three methods for defining groups, that is, three types of group construction or representation. For another, see Section 12.1. They are defined using (a) permutations, (b) matrices, or (c) generators and relations, that is, presentations, and they all have important roles to play in the theory. We begin by describing the basic properties of permutations, and the symmetric and alternating groups. Historically, the development of the theory started with them and A. Cayley (1821Ð1895) in 1850 showed that every group can be represented as a permutation group (Theorem 4.7). And it is for this reason that some authors describe group theory as ‘the science of symmetry’; see Weyl (1952). Next we give the basic properties of matrix groups, many of the more ‘interesting’ groups in the theory, especially many simple groups, arise first as matrix groups; see Chapter 12. Lastly, we give the formal definition of a group presentation which will be further discussed and verified in Web Section 4.7.

3.1 Permutations

We begin by developing the basic properties of permutations. Remember that we always read from left to right. Most of the work in this section ﬁrst appeared in print in a series of papers published in the 1840s by the French mathematician A. Cauchy (1789Ð1857); see Section 6.1. The notation introduced below is also due to him.

Deﬁnition 3.1 A permutation σ onasetX is a bijection of X to itself.

As we shall normally be using finite sets X, it is convenient, but not essential, to take X ={1, 2,...,n} when o(X) = n. Apart from their natural ordering, no arithmetical properties of the integers 1 to n are used, they are just easily recognised labels for the elements of a set with n elements. We use two notations. First we have the ‘matrix’ form: 12... n σ = , (3.1) a1 a2 ... an where ai ∈ X and iσ = ai ,fori = 1,...,n; see the note on ‘left and right’ on page 68. Each element in the second row of (3.1) is the result of applying the permutation σ to the element in the first row directly above it: i → ai , i = 1,...,n, so no two elements in the second row are equal (some authors just print the second row taking our first row as read). The order of the columns is unimportant, but we usually write them as in (3.1). Using the fact that composition of two bijections is a bijection (Appendix A), we define the ‘product’ of two permutations by

Deﬁnition 3.2 Let σ and τ be permutations on the same set X, where, for i ∈ X, iσ = ai and iτ = bi .Theproduct στ is given by = = ∈ i(στ) (iσ )τ bai for all i X. 3.1 Permutations 43

Using the matrix form (3.1) above, we can rewrite this as 12... n 12... n στ = a1 a2 ... an b1 b2 ... bn = 12... n a1 a2 ... an a1 a2 ... an ba1 ba2 ... ban 12... n = , ba1 ba2 ... ban where the second matrix in the second line is the same as the second matrix in the ﬁrst line except that its columns have been permuted by σ . This does not affect the result.

In the next section, we show that this product generates a number of new groups. The neutral element is the permutation that moves no element of X (the identity map ι on X), and the inverse of a permutation is its reverse, that is, if σ is given by (3.1), then − a a ... a σ 1 = 1 2 n . 12... n

The second notation for permutations uses cycles. We begin with an example. Let 123456789 σ = . 1 731852694

This permutation maps 1 → 7, 7 → 6, 6 → 2, 2 → 3 and 3 → 1 so forming a cycle with ﬁve entries (1, 7, 6, 2, 3). Alternatively, we can write

= 2 = 3 = 4 = 5 = 1σ1 7, 1σ1 6, 1σ1 2, 1σ1 3, and 1σ1 1.

As the symbol 4 has not been used so far, we can start again: σ1 maps 4 → 8, 8 → 9 and 9 → 4, giving another cycle with three entries (4, 8, 9) in this case, which is disjoint from the ﬁrst cycle. Finally, 5 is the only symbol in X which has not so far been used, and 5σ1 = 5 giving a third cycle with only one element. So we can treat σ1 as the ‘product’ of these three cycles, that is, we write

σ1 = (1, 7, 6, 2, 3)(4, 8, 9)(5)

(although sometimes single cycles, like (5), are taken as read and not printed). Note that the cycles are disjoint from one another, and the order in which they are written does not affect the ﬁnal outcome. This example is typical as we show below. 44 3 Group Construction and Representation

Deﬁnition 3.3 Let σ be a permutation on the ﬁnite set X and let x ∈ X. The ordered k-tuple x,xσ,xσ2,...,xσk−1 , where k is the smallest positive integer with the property x = xσk is called the cycle of length k,orthek-cycle, containing x. The integer k is called the length of the cycle.2 Some authors use the term ‘transposition’ for a 2-cycle, say (x, y),thatis,a permutation which maps x → y and y → x.

Notes We use the same notation for an ordered k-tuple and a cycle, no confusion should arise. No two elements of a cycle are equal and 1 ≤ k ≤ o(X). Also compared with our ﬁrst ‘matrix’ notation, cycle maps are horizontal left to right except that the last entry is mapped to the ﬁrst, see footnote.

The following result gives the essential facts about cycles.

Theorem 3.4 Suppose σ is a permutation of the set X ={1, 2,...,n}. −1 (i) If τ is the cycle (a1,a2,...,an), then τ = (an,an−1,...,a1). (ii) The permutation σ can be expressed as a product of cycles τ1,τ2,...,τk, where k ≥ 1. They are disjoint and commute in pairs. (iii) The representation of σ given in (ii) is unique except for the order in which the cycles τi appear in the product.

Proof (i) This follows immediately from the deﬁnition. (ii) We repeat the argument given in the example above. The sequence

1, 1σ, 1σ 2,...

forms a cycle C1 of length k1, where k1 is the least positive integer satisfying k 1σ 1 = 1. If C1 = X, the result follows, for then σ forms a single cycle. If not, let a1 be the least positive integer in X not used in C1, and consider the 2 cycle C2 = (a1,a1σ,a1σ ,...) of length k2, where k2 is deﬁned in a similar way to k1.Now

C1 ∩ C2 =∅. (3.2) r s For if not, positive integers m and n exist satisfying 1σ = a1σ , which gives r−s a1 = 1σ contrary to our assumption that a1 ∈/ C1. We can continue this process forming C3,C4,... until all of X is used up, and then σ = C1C2 ···. By (3.2), each Ci is disjoint from the other cycles, and they commute for the same reason. (iii) The result clearly holds for 1-cycles. Suppose σ = τ1 ···τr = ν1 ···νs where τ1,...,τr (ν1,...,νs , respectively) are disjoint cycles each of which

2For ease of typesetting, cycles are printed linearly, but perhaps they should be printed in a circle as this would more truly represent them. 3.1 Permutations 45

moves at least two entries. Let 1 ≤ i ≤ n, and suppose both τk and νl move = = t = t = t i,soiτk iσ iνl . Then iτk iσ iνl for all t by assumption. Us- ing Problem 3.4(iii), this shows that τk = νl , and so τ1 ···τk−1τk+1 ···τr = ν1 ···νl−1νl+1 ···νs . We can repeat this argument, so use induction.

Note that in this proof k1 + k2 +···=n = o(X). We need to prove the following two results before we can introduce the next topic—even and odd permutations.

Lemma 3.5 Every permutation on a ﬁnite set can be expressed as a product of 2-cycles.

Proof By Theorem 3.4, every permutation σ can be expressed as a product of cycles. Hence we need to show that every cycle equals a product of 2-cycles. But this follows by relabelling from the identity

(1, 2,...,n)= (1, 2)(1, 3) ···(1,n),

which the reader should check noting we read from left to right.

We can extend this result to: Every permutation on X ={1, 2,...,n} can be expressed in terms of the 2-cycles (1, 2), (1, 3),...,(1,n); see Problem 3.1. Cyclic structure is preserved by conjugation, see Deﬁnition 2.28.Wehave

Theorem 3.6 Suppose σ and τ are permutations on X ={1, 2,...,n}. The permutations σ and τ are conjugate (a permutation α on X exists satisfying τ = α−1σα) if and only if σ and τ have the same cyclic structure, in which case τ can be obtained by applying α to the symbols of σ .

First, we consider an example. Let X ={1,...,6}, and let

σ = (1, 5)(3)(2, 6, 4), τ = (2, 3)(6)(4, 5, 1) and α = (1, 2, 4)(3, 6, 5).

We have 2α−1 = 1, 1σ = 5 and 5α = 3, and so 2α−1σα = 3. This agrees with τ which also maps 2 to 3. Repeating this for the other members of X shows that α−1σα= τ .Alsoas1α = 2,...,6α = 5, we have

(1α, 5α)(3α)(2α, 6α, 4α) = (2, 3)(6)(4, 5, 1) = τ, that is, if we replace the entries in σ by their images under α, we obtain τ . Con- versely, note that if we construct a ‘permutation matrix’ whose top row is the entries of σ , and whose bottom row is the entries of the cycle τ , we obtain α, viz.: 153264 α = , 236451 46 3 Group Construction and Representation which is a matrix form of α. This only works because σ and τ have the same cyclic structure—a product of a 2-cycle, a 1-cycle and a 3-cycle.

Proof Suppose ﬁrst σ and τ have the same cyclic structure with cycle lengths corresponding:

σ = (...)···(...,l,m,...)···(...), τ = (. . .) ···(...,l ,m ,...)···(...).

Form α (using its ‘matrix’ form, see (3.1)) by taking the entries of σ as its top row and the entries of τ as its bottom row: ... l m ... α = . (3.3) ... l m ...

Now as lσ = m, lα = l so l α−1 = l, and mα = m ,wehave

− mα = m so lσα = m , and so l α 1σα= m .

But l τ = m , therefore we can deduce τ = α−1σα, that is, σ and τ are conjugate, because the above calculation can be applied to all corresponding consecutive pairs of entries in σ and τ (if l is the last entry in a cycle then m is the ﬁrst; see footnote on page 44). For the converse, suppose σ has the form

σ = (...)···(...,j,k,...)···(...),

as above. Further, suppose jα = r and kα = s, then as jσ = k we have

− kα = s so jσα = s, and so rα 1σα= s.

Now as α and σ are permutations (bijections on X), so is α−1σα. Therefore, if we deﬁne τ by τ = α−1σα, the conjugate of σ by α, then it has the same cyclic structure as σ because the replacements l → r and m → s are themselves bijective, and the cyclic structure of σ is unaltered by this procedure.

Returning to the example on page 45, we note that to obtain the given α we had to write σ and τ in the ‘right’ way. The cycles (4, 5, 1), (1, 4, 5) and (5, 1, 4) are identical, and so a number of different α can be constructed using the process given in the proof of the theorem. This is to be expected because there is likely to be a number of different solutions α to the equation ατ = σα. For example, we can use 153264 α = = (1, 2)(3, 6, 4, 5). 236145 3.1 Permutations 47

The reader should check that this new α also gives the correct answer, and write out the remaining possible solutions (Problem 3.2).

Even and Odd Permutations

A permutation is even (odd) if it can be expressed as a product of an even (odd, respectively) number of 2-cycles. This is not a deﬁnition as it stands because there are many ways of expressing a permutation as a product of 2-cycles (Lemma 3.5). For example, the permutation (5, 6)(3, 4) equals, amongst others,

(1, 2)(4, 3)(5, 6)(2, 1), (2, 3)(2, 4)(2, 3)(6, 5) and (5, 1)(2, 3)(2, 4)(1, 6)(1, 5)(3, 2), and so they all represent the same permutation. But in each case the number of 2-cycles is even; this is typical of the general situation as we show now. Suppose o(X) = n and we are considering permutations on X. We introduce the following polynomial f which we use to ‘codify’ all possible 2-cycles. It is given by

f(x1,...,xn) = (xj − xi). (3.4) 1≤i

fσ(x1,...,xn) = f(x1σ ,...,xnσ ) = (xjσ − xiσ ). 1≤i

Clearly, as σ is a permutation on X, each factor xj − xi of f in (3.4) occurs as a factor of fσ and vice versa, but the sign of this factor may be altered by σ . Hence, for all σ we have fσ =±f , and we use this fact to deﬁne the terms even and odd.

Deﬁnition 3.7 (i) Using the notation set out above, we say that the permutation σ is even if fσ = f , and odd if fσ =−f . (ii) The sign,sgn(σ ), of the permutation σ is given by

1ifσ is even, sgn(σ ) = −1ifσ is odd.

We show now that our two ‘deﬁnitions’ agree, and begin by proving

Lemma 3.8 If the permutation σ is a single 2-cycle, then sgn(σ ) =−1, that is fσ =−f where f is given by (3.4). 48 3 Group Construction and Representation

Proof Suppose σ is the 2-cycle (i, j) where 1 ≤ i

(xk − xi)σ =−(xj − xk) and (xj − xk)σ =−(xk − xi).

These sign changes occur in pairs and so when combined they have no overall effect, the result follows.

Theorem 3.9 (i) If σ,τ are permutations on X, then sgn(σ τ) = sgn(σ ) sgn(τ).

k (ii) If σ = τ1 ···τk and each τi is a 2-cycle, then sgn(σ ) = (−1) .

Part (ii) shows that our two deﬁnitions of even and odd agree.

Proof (i) For i ∈ X,wehave(iσ )τ = i(στ) by associativity of composition (Appendix A), hence f (σ τ)(x1,...,xn) = xj(στ) − xi(στ) 1≤i

3.2 Permutation Groups

In the previous section, we studied properties of permutations, here we use these properties to construct a number of groups. Historically, these developments had a considerable inﬂuence on the progress of the theory as a whole; in the next chapter (Theorem 4.7), we show that every group can be treated as a group of permutations—a result due to Cayley. Note that there is nothing unique about permutations, in Problem 4.17 we show that every ﬁnite group can also be treated as a matrix group in many different ways; see also Web Section 4.7. First, we prove that the collection of all permutations on a set forms a group.

Theorem 3.10 Suppose X is a set. The collection of all permutations on X, with composition as the operation, forms a group. 3.2 Permutation Groups 49

Proof A permutation is a bijection on X. The facts that composition of two bijections is a bijection, and composition of maps is associative, are proved in Appendix A. The neutral element is the permutation which moves no elements of X (the identity map ι, page 281), and the inverse of a permutation σ is its reverse: If a,b ∈ X and aσ = b, then bσ −1 = a; see page 43. This proves the theorem.

The group of all permutations on X is denoted by SX.Ifo(X) = n (where we usually take X ={1, 2,...,n}), the group is denoted by Sn and called the symmetric group on n symbols. These groups have the following basic properties:

(a) o(Sn) = n!. There are n! distinct bijections of an n-element set to itself. (b) If m ≤ n, then Sm ≤ Sn.InSn, consider all permutations which keep the same n − m elements fixed. For instance, S4 ≤ S5; in fact, S5 contains five copies of S4; see Problem 4.20(ii). (c) S1 = e , S2 is cyclic, and Sn is not Abelian if n>2; for instance, the cycles (1, 2) and (1, 2, 3) do not commute. (d) By Lemma 3.5, Sn is generated by its 2-cycles, it can also be generated by a two element set; see Problem 3.1. (e) By Theorem 3.6, two elements σ,τ ∈ Sn are conjugate in Sn (Definition 2.28)if and only if they have the same cyclic structure—a result we use several times.

The subset (subgroup, see Theorem 3.11 below) of even permutations in Sn given by Deﬁnition 3.7 forms an important subgroup of Sn;forn>4, this provides our ﬁrst example of a class of non-Abelian simple groups. We begin with

Theorem 3.11 For n>1, the set of even permutations in Sn forms a normal subgroup of Sn with order n!/2.

Proof A permutation is even if it can be expressed as an even number of 2-cycles (Theorem 3.9). Clearly, the identity permutation is even. Also the product of two permutations each a product of an even number of 2-cycles is itself a product of an even number of 2-cycles, and the inverse of a product of 2-cycles is the product written in reverse order. Hence, by Theorem 2.13, the set of even permutations forms a subgroup of Sn. It is normal by Theo- rems 2.29 and 3.6, as this last result shows that conjugation does not alter the cyclic structure of a permutation. Exactly half of all permutations in Sn are even, hence the order of the subgroup is n!/2. We prove this as follows. Deﬁne a map θ from the set of even permutations to the set of odd permutations by

σθ = (1, 2)σ where σ is even.

This is a bijection, for if τ is odd, then (1, 2)τ is even and (1, 2)τθ = (1, 2)(1, 2)τ = τ . Hence θ is surjective. A surjective map on a ﬁnite set onto itself is injective (Appendix A). The result follows. 50 3 Group Construction and Representation

This last property extends to all subgroups of Sn, see the example on page 77 and Problem 3.7(ii).

Alternating Groups

The subgroup of even permutations of Sn given by Theorem 3.11 is called the alternating group on n symbols, it is denoted by An. Note that o(A1) = o(A2) = 1, and o(A3) = 3asA3 contains just two 3-cycles [(1, 2, 3) and (1, 3, 2)] and the identity permutation. A4 is the only non-Abelian alternating group that contains a (proper non-neutral) normal subgroup; see Problem 3.10 and page 158. 3 For n>4, the groups An are simple, and they form our second inﬁnite collection of simple groups. We give an elementary proof of this simplicity result now; see comments at the end of the proof. The ﬁrst step is to show that An is generated by its 3-cycles in the same way that Sn is generated by its 2-cycles (Lemma 3.5).

Theorem 3.12 The group An is generated by its 3-cycles, provided n>2.

Proof By Theorem 3.9, every element in An is equal to a product of an even number of 2-cycles. We prove the result by showing that all products of two 2-cycles are products of 3-cycles. Suppose 1 ≤ i, j, k, l ≤ n, and they are distinct. Then we have (i, j)(i, j) = e, (i, j)(i, k) = (i,j,k) and (i, j)(k, l) = (i,l,j)(j,k,l); the result follows.

We come now to the main simplicity proof. We begin by assuming the contrary, that is, a proper subgroup K exists satisfying e =K An, then we show (a) if K contains a 3-cycle, then K contains all 3-cycles, and so by Theorem 3.12, K = An; and (b) K contains a 3-cycle if n>4.

These two propositions prove the theorem because they show that An does not contain a proper non-neutral normal subgroup. First, we prove (a).

Lemma 3.13 If K An and K contains a 3-cycle, then K = An.

Proof As K An, K contains conjugates of all of its elements by Theo- rem 2.29. Hence, by Theorem 3.12, we need to show that 3-cycles are conjugate in An. By Theorem 3.6, all 3-cycles are conjugate in Sn, but we cannot apply this directly because the conjugating element relating two 3-cycles may not belong to An; we must prove that all 3-cycles are conjugate by even elements. We show this directly using elementary permutation arguments, later

3For n = 5 this result is effectively due to Abel, it was a corollary of his work on the non-solubility of the quintic. 3.2 Permutation Groups 51

(Problem 5.25) we give another proof of this result which ‘sheds more light’ on the underlying structure. We consider A5 ﬁrst. Two distinct 3-cycles will overlap by one or two elements. Suppose there is a single overlap then, relabelling if necessary, we can take the 3-cycles in the form

(1, 2, 3) and (1, 4, 5).

We have (2, 4)(3, 5) ∈ A5 and

(2, 4)(3, 5)(1, 2, 3)(3, 5)(2, 4) = (1, 4, 5).

Similarly, if there is a double overlap, that is, the cycles are of the type (1, 2, 3) and (1, 2, 4),or(1, 2, 3) and (2, 1, 4), then (1, 4, 3), (3, 4, 5) ∈ A5 and

(5, 4, 3)(1, 2, 3)(3, 4, 5) = (1, 2, 4) (3, 4, 1)(1, 2, 3)(1, 4, 3) = (2, 1, 4).

The result follows for A5 by permuting the symbols 1,...,5. Secondly, we use this to prove the general result. If n>5, then An contains copies of A5 obtained by considering only those permutations that move the symbols of a ﬁxed 5-element subset of {1,...,n}. In one of these copies of A5, moving the symbols 1, 2, 3,l,∗ only, we have (1, 2, 3) is conjugate to (1, 2,l), and so they are also conjugate in An. In a second copy, moving the symbols 1, 2,j,k,lonly, we have (1, 2,l)is conjugate to (j,k,l)in An. Hence (1, 2, 3) is conjugate to (j,k,l)in An, and the lemma is proved when n>4. (It is also true when n = 3 or 4, the reader should check this.)

Theorem 3.14 If n>4, then the group An is simple.

Proof We use a similar method to that given in the previous proof. Suppose K An and e =K. As noted above ((b) on page 50), we need to show that K contains a 3-cycle, the theorem then follows by Lemma 3.13.Letτ ( = e) be an element of K which moves the least number of symbols in the underlying set {1,...,n}. We show that τ is a 3-cycle. It cannot be a 2-cycle because 2-cycles are odd, hence if it is not a 3-cycle, it must move at least four symbols. It cannot be a 4-cycle because 4-cycles are odd. Hence it satisﬁes one of: (i) τ is a product of an even number of disjoint 2-cycles, (ii) τ is a product of a cycle of length at least 3 and further disjoint cycle(s).

Therefore, τ is one of the following types τ1 or τ2, where σi is a cycle or a product of cycles whose entries are disjoint from those of its predecessors:

(i) τ1 = (1, 2)(3, 4)σ1 (ii) τ2 = (1, 2, 3,...)(4, 5,...)σ2. 52 3 Group Construction and Representation

In each case σi may be absent, and in (ii) the ﬁrst cycle in τ2 has length at least 3, also the second cycle may have length 2 or may be longer, but we write (one of) the longest cycles ﬁrst. We use τi to construct a new permutation which is not e and which moves fewer symbols than τi , this contradicts our supposition and so proves the theorem. In both cases, this new permutation is

[ ]= −1 −1 = τi,α τi α τiα where α (3, 4, 5).

We show [τi,α] moves fewer symbols than τi . Note that as τi ∈ K An,all conjugates of τi by elements of An belong to K, and so [τi,α]∈K. (i) Applying the permutations that make up [τ1,α] in turn, we have:

beginning with the base set 1 2 3 4 5 ..., −1 ∗ applying τ1 we obtain 2 1 4 3 ..., applying α−1 we obtain 2 1 3 5 ∗ ...,

applying τ1 we obtain 1 2 4 ∗∗..., applying α we obtain 1 2 5 ∗∗....

So if we apply [τ1,α] to our base set we obtain a permutation which ﬁxes 1 and 2, and maps 3 to 5, that is, we have in K a non-neutral permutation which moves fewer symbols than τ1, contradicting our assumption. (ii) If we apply the permutation [τ2,α] to our base set {1, 2, 3, 4, 5,...} we obtain a permutation which ﬁxes 2 and maps 3 to 4, and so again we have a contradiction; the reader should check this. In both cases, we have constructed in K a non-neutral permutation moving fewer symbols than those moved by τ . Therefore, our assumption is false, τ is a 3-cycle, and the theorem follows.

There are many proofs of this result in the literature. The proof given above has the advantage that it uses a minimal amount of ‘apparatus’, and so it can be presented here. Further proofs of this result are given in Problems 3.16 (using conjugation), 5.25 (using centralisers), and 6.16 (using the Sylow theory) and in Web Sections 3.6 and 14.1. Suzuki (1982, page 295) gives a proof that uses Bertrand’s 2 postulate! He shows that o(An) = o(K) if K An, and then he applies the postulate (which states that for all positive integers m aprimep can be found lying between m and 2m). Up to isomorphism, A5 is the only non-Abelian simple group with order less than 168, see Problems 6.15 and 6.17, and Chapter 12.

3.3 Matrix Groups

Matrix algebra provides a wide range of group examples. Given a field F and a positive integer n, we consider the set of all n × n non-singular matrices with entries in F .(Ann × n matrix A is non-singular if and only if another n × n matrix B 3.3 Matrix Groups 53 exists satisfying AB = BA = In. A standard theorem of linear algebra states that A is non-singular if and only if det A = 0 where det A denotes the determinant of A.) This set of matrices forms a group called the general linear group over F , and it is denoted by GLn(F ). The neutral element is the n × n identity matrix In, and inverses exist by definition. The normal subgroup (Theorem 3.15 below) of matrices with determinant 1 is denoted by SLn(F ) and called the n × n special linear group over F . We shall mainly be concerned with the case when F is finite, see Section 12.2. Up to isomorphism, unique finite fields exist of order pm for each prime number p and positive integer m; they are usually called Galois fields, and denoted by Fpm , after their discoverer É. Galois;4 see page 229. The general linear group defined over m m a field with p elements, that is, GLn(Fpm ), is denoted by GLn(q) where q = p . Similarly, we write SLn(q) for SLn(Fpm ). The basic properties of these groups are given by

Theorem 3.15 Suppose F is a ﬁeld and n ≥ 1.

(i) The set of matrices GLn(F ) with matrix multiplication forms a group. (ii) SLn(F ) GLn(F ). (iiia) If q = pm, where p is prime and m>0, then n(n−1)/2 n n−1 o GLn(q) = q q − 1 q − 1 ···(q − 1).

(iiib) If n>1, o(SLn(q)) = o(GLn(q))/(q − 1), and o(SL1(q)) = 1.

Proof (i) and (ii) Both of these follow from the fact that det AB = det A det B −1 for all A,B ∈ GLn(F ), and its corollaries det A = 1/ det A, and det In = 1. (iiia) Let A ∈ GLn(q).AsA is non-singular, every sequence of n elements n of Fq is a possible top row of A except {0, 0,...,0}, that is, there are q − 1 possible top rows for A.AgainasA is non-singular, a possible second row of A is a sequence of n elements of Fq which is not a linear multiple of the first row. There are q multiples of the first row (this includes the row {0, 0,...,0}), so qn − q = q(qn−1 − 1) second rows of A are possible. We can continue this process, the third row must not be a linear combination of the first two rows, and so on. The result follows by collecting terms. (iiib) This is a consequence of (ii) and (iiia) as the number of non-zero elements of Fq is q −1, see page 76, or we can use the following argument. In the last stage of the procedure given in (iiia), there are qn − qn−1 = qn−1(q − 1) choices for the last row of the matrix. So there are qn−1 choices if we also stipulate that the determinant of the matrix in question has a particular value. This gives (iiib) if we choose this value to be 1; the reader should check this.

For example, we have o(GL3(2)) = 168 and o(SL2(3)) = 24.

4The fact that all finite fields have this form was first proved by E.H. Moore in 1893. 54 3 Group Construction and Representation

The groups SLn(q) give rise to our third class of simple groups. If SLn(q) is ‘factored’ by its centre then the resulting linear group which is denoted by Ln(q) is simple, except when n = 2 and q = 2 or 3. The operation of forming a factor group will be discussed in the next chapter, and a simplicity proof will be given in Chapter 12. This result implies that SLn(q) is ‘nearly’ simple, its only (proper non- neutral) normal subgroup being its centre which has order (q − 1,n). In some cases, SLn(q) is itself simple (when its centre has order 1), examples are SL2(4) (which is isomorphic to A5, see Problem 6.16) and SL3(3) (Section 12.2).

Subgroups of GLn(q)

The group GLn(q) has a number of important subgroups, one is SLn(q) as discussed above. Another type is the subgroup of permutation matrices discussed in Problem 3.12. A third type is the subgroup of ‘upper triangular’ matrices. (For a full list, see Dickson 2003, or Kleidman and Liebeck 1990.) A matrix A ∈ GLn(q) is called upper triangular if every entry in A below the main diagonal is zero, this subgroup is denoted by UTn(q). Lower triangular matrices can also be introduced. Note that each diagonal entry of a non-singular upper triangular matrix is non-zero because the determinant of this matrix equals the product of its diagonal elements. We have

Lemma 3.16 UTn(q) ≤ GLn(q).

Proof Clearly, In ∈ UTn(q), and the product of two upper triangular matrices is itself upper triangular. Hence, by Theorem 2.13, we need to show that the inverses are also upper triangular. Let A = (aij ) ∈ UTn(q) (where aij is the (i, j)th entry in the matrix A) and so aij = 0ifj

sij = aiibij +···+ainbnj ,

with n − (i − 1) summands as ai1 = ··· = ai(i−1) = 0. If j

s(n−1)j = a(n−1)(n−1)b(n−1)j + a(n−1)nbnj = 0.

This gives b(n−1)j = 0 when j

Suppose A ∈ GLn(q). We can ﬁnd U1,U2 ∈ UTn(q) and a permutation matrix P (Problem 3.12) to satisfy

A = U1PU2. This result is known as the Bruhat Decomposition Theorem, and it has a number of useful applications. We shall give a simple proof in the case n = 2, the proof of the 3.4 Group Presentation 55 = ab = general case follows similar lines. Let A cd .Ifc 0, then the result follows by putting U1 = A and P = U2 = I2,the2 × 2 identity matrix. So we may assume c = P = 01 U = x1 y1 U = x2 y2 that 0. In this case, we let 10 , 1 01 and 2 01 . Then, if U1PU2 = A we have x y 01 x y y x y y + x ab 1 1 2 2 = 1 2 1 2 1 = . 01 10 01 x2 y2 cd

−1 −1 This gives x2 = c = 0, y2 = d, y1 = ac and x1 = b − dac = 0, as det A = 0. Now this deﬁnes both U1 and U2. × transvection 1 r 10 r = A2 2 is a matrix of the form 01 or r 1 where 0; we shall consider these matrices in more detail in Chapter 12. Here we use them to derive a consequence of Bruhat’s result as follows.

Theorem 3.17 The group GL2(q) is generated by its diagonal matrices and its transvections.

Proof By Bruhat’s Theorem we need to show that we can construct the upper triangular and permutation matrices as products of diagonal matrices and transvections. We have a 0 1 ba−1 ab = , 0 c 01 0 c 11 10 01 = , and 01 −11 −11 01 −11 01 = , −11 01 10

which proves the result as there are only two permutation matrices in the 2-dimensional case.

3.4 Group Presentation

The American mathematician W. von Dyck (1856Ð1934) in about 1880 introduced the ‘presentation’ of a group, as an example he derived the presentation of S4 given in Problem 3.18. This work has led to a branch of the theory called Combinatorial Group Theory, we shall only give the basic ideas and definitions, the interested reader should consult Lyndon and Schupp (1976) or Chandler and Magnus (1982). In the second section of Chapter 2, we introduced briefly a method for defining groups using so-called generators and relations, here and in Web Section 4.7 we give the formal definitions and proofs. The method is important in the theory because many groups are best defined in these terms, see, for example, the group E discussed in Section 8.3. Although it is a representation, it is the nearest ‘approach’ to an abstract definition of the group. 56 3 Group Construction and Representation

Suppose we are given an alphabet or list A of letters or symbols which we usually assume to be ﬁnite (but this is not essential):

a1,a2,...,b1,...,c1,....

A word is deﬁned as a ﬁnite sequence of letters written using concatenation, for example,

b2,c3c3b1a2c1, and a1a1a1a1 are words. In this section, we use lower case characters at the beginning of the alphabet a,b,c,... for letters, and lower case characters at the end of the alphabet x,y,z,... for words.Ifthewordx has the form x = y1zy2, then z is called a sub- word of x (as are both y1 and y2). The operation is as suggested above, that is, concatenation. For example, if x = aiaj bk and y = bkclc1am, then

xy = aiaj bkbkclc1am.

It follows immediately that the system of all words on a ﬁxed alphabet A with the operation of concatenation forms a semigroup. To construct a group we need to bring the neutral element and inverses into the system, and we do this as follows. The word containing no letters from A will act as the neutral element. Mainly for typographical reasons, it is necessary to introduce a symbol for this element, and so as previously we let e stand for it where e/∈ A.Note that the symbol e and the blank symbol are synonymous. For the inverse operation to apply, the alphabets have the form A ∪ A , where A ∩ A =∅and there exists a bijection between A and A . So the alphabet has the structure:

a1,a1,a2,a2,...,b1,b1,...,c1,c1,..., and the given prime bijection between A and A satisﬁes (a ) = a, for all letters a ∈ A and a ∈ A . The basic idea here is that we want a to act as the inverse of a, et cetera. We deﬁne the set of reduced words on A by

Deﬁnition 3.18 Awordx on the alphabet A (it is assumed that the second alphabet A is also present; see above) is called a reduced word if (a) it contains no pairs of consecutive letters (symbols) of the form aiai,aiai,bj bj ,..., that is, it contains no letter, with or without a prime , immediately followed by its image under the -map, and (b) all blank symbols have been removed except that if no alphabet letters remain, then a single blank symbol should remain.

For example,

c3,a1b2,a1a1a1,e are reduced, and

a3a3,b2e,b1c1c1,a2a2a2a2 are not reduced. 3.4 Group Presentation 57

We now adapt the semigroup concatenation operation to construct a group operation. Suppose x and y are reduced words. To form their ‘product’, which we call the reduced concatenation of x and y, we ﬁrst construct the semigroup concatenation xy of x and y, and then we remove (that is, replace by the empty word) all pairs of consecutive symbols of the form aiai ,oraiai ,orbj bj , et cetera, that are formed = by the concatenation, or by previous removals. For example, if x a1a1b2c1 and = y c1b2a1, then = = = = ; xy a1a1b2c1c1b2a1 a1a1b2b2a1 a1a1a1 a1 = = and yx c1b2a1a1a1b2c1 c1b2a1b2c1. (This construction is a generalisation of one that will be familiar to the reader. In the group of integers Z, we have, for instance, 3 = 4 − 1 = 5 − 2 =···=(n + 3) − n. Note (n + 3) − n = 3 + (n − n) = 3 for all n ∈ Z, that is, each time we encounter n − n we replace it by 0, and delete it if other symbols are present. This is exactly mirrored in the general case described above.) This system forms a group. Two words are equal if each can be obtained from the other by the insertion and/or deletion of pairs of consecutive symbols of the form aa or a a. To be more precise this procedure deﬁnes an equivalence relation on the set of words, and the group elements are the corresponding equivalence classes, see Web Section 4.7 (In our example using the group Z given above, we associate 3 with the set of differences {(n + 3) − n : n ∈ Z}.) The set of words has a neutral element and is closed under the inverse operation, hence we need to establish associativity. We have

Theorem 3.19 The system of reduced words on a ﬁxed alphabet A (that is, A ∪ A , see page 56), with the operation of reduced concatenation deﬁned above, forms a group.

We shall not give a proof of this result here. This is best done once we have proved the First Isomorphism Theorem which is given in Chapter 4;seeWeb Sec- tion 4.7. It is also possible to give an ad hoc proof which splits the result into a number of particular cases, but it is somewhat unsatisfactory in that it does not illustrate the underlying structure, and it is not easy to be sure that all cases have been considered. Reader, try it. The group defined by the above theorem is called the free group on the alphabet A. It is called free because there are no constraints on the words in the group other than those needed to form the group. It is necessarily infinite, for if a ∈ A then a,aa,aaa,... all belong to the group, and they are distinct. A range of new groups can be defined by introducing more constraints. Consider the following example. Let A ={a}, that is, the alphabet consists of the two letters a and a where aa = a a = e, the empty word, and let G be the free group on A. Then the (reduced) elements are ...,a a ,a ,e,a,aa,aaa,..., 58 3 Group Construction and Representation and G is called the infinite cyclic group, it is denoted by Z. We can introduce a relation in the form an = e where n is a positive integer and an stands for aa ···a with n copies of a. It is easily seen that the elements of the new structure are e,a,a2,...,an−1, and that it forms a group with n elements. It is called the cyclic group of order n and is denoted by Cn. We can treat the elements of Cn in the same way as those of the infinite cyclic group except each time we encounter an we delete it (replace it bytheemptyword)aswedoforaa and a a. Further cyclic group properties are given in Section 4.3. The procedure illustrated in this example can be applied to all free groups. Sup- pose A is an alphabet and R is a set of words on A (that is, on A ∪ A ), then we can form the structure (group) H called a presentation. It is denoted by H = A | R , (3.5) and consists of the free group on A with the constraints (relations) x = e, for all x ∈ R.IfR ={x1,...,xk}, we usually write x1 =···=xk = e for R in this presentation. The elements of A are called the generators of H ,forx ∈ R the equation x = e is called a relation of H , and this method of defining H is called a presentation of H . (Some authors use the term relator for a word x in R.) To study H we work in the free group on A, and each time we encounter a word in R we replace it by the empty word just as we do for aiai,aiai,... in the free group. In the example n n above, we have Cn = a | a = e and a is replaced by the empty word each time it occurs. For all A and R the structure H forms a group, we shall prove this in Web Section 4.7 using the Isomorphism Theorems. We show that H can be defined as a ‘factor group’ of the free group on A. We also show that all groups can be treated in a similar manner. It can be difficult, and in some cases impossible, to determine the properties of H (see (3.5)), its order, or even if it is finite or infinite. For a general discussion of this and related considerations, including the Word Problem for Groups, see Rotman (1994). One difficulty is of the following type: Given a collection of relations R and a generator a, we might be able to deduce both am = e and an = e where (m, n) = 1, this would give a = e (use the Euclidean algorithm) and so a would be redundant. In the ‘worst’ case, the whole construction could collapse to the neutral group. To avoid this, and similar problems, in most cases it is necessary to find another representation of the group in question, as a matrix, permutation, or similar group, and to construct a bijection between the corresponding elements. We illustrate this in the following two examples. Todd and Coxeter have devised a method called coset enumeration which can be used to determine the structure of a group given by a presentation; see Coxeter and Moser (1984), Chapter 2.

n 2 2 Example 1 (Dihedral group Dn)LetA1 ={a,b} and R1 ={a ,b ,(ab) }. Then n 2 a,b | a = b = abab = e gives a presentation of the dihedral group Dn;see 3.5 Problems 59

Section 2.2. The order is 2n because it can easily be shown that every element of the free group on {a,b} can be reduced to a member of the set

e,a,a2,...,an−1,b,ab,...,an−1b using the relations in R1. There is no collapse, see above, because our ﬁrst representation of this group was as the symmetry group of the regular polygon with n sides and 2n symmetries. The group can also be generated by two involutions, see Problem 3.20.

n −2 −1 Example 2 (Dicyclic Group Qn) Again let A1={a,b}, and let R2={a b ,abab }, n 2 2 and so Qn = a,b | a = b = (ab) . We show ﬁrst that the relation

a2n = e (3.6) is a consequence of the relations in R2; note we are assuming that the structure a,b | R2 is a group, this is proved in Web Section 4.7. Hence we can apply n 2 basic group properties. Assume ﬁrst that n is even. The relations R2 give a = b and b = aba, and so we have, replacing b by aba several times:

an = b2 = (aba)(aba) = a2ba4ba2 =···=an/2banban/2 = an/2b4an/2 = a3n, and so in this case (3.6) follows by cancellation; we leave it as an exercise for the reader to derive the remaining case. Now as in the ﬁrst example we can show that a member of the group is of the form at bu where 0 ≤ t<2n and 0 ≤ u ≤ 1, and so the group has order 4n. Matrix representations of these groups are given in Prob- lems 3.13 and 3.22.Weusethetermquaternion group for the dicyclic group Q2, see Section 6.1. Also the term generalised quaternion group is used for the group (n+2) Q2n (with order 2 ) when n = 2, 3,...; see Problem 3.22.

3.5 Problems

Problem 3.1 Show that Sn can be generated by each of the following sets: (i) {(1, 2), (1, 3), . . . , (1,n)}; (ii) {(1, 2), (2, 3),...,(n− 1,n)}; (iii) {(1, 2), (1, 2,...,n)}.

As every finite group G can be treated as a subgroup of Sm for some suitably chosen m where m ≤ o(G) (Cayley’s Theorem, page 72), (iii) shows that every finite group is a subgroup of a group with two generators, or to put this another way, increasing the number of generators (above two) does not necessarily increase the complexity of the group. Also all finite non-cyclic simple groups have two element generating sets; see Cameron (1999).

(iv) Show that An,forn>4, is generated by its involutions. Note that this property holds for all non-Abelian simple groups (Problem 2.28). 60 3 Group Construction and Representation

Problem 3.2 If σ = (1, 2, 3)(4, 5, 6) and τ = (1, 5, 6)(2, 3, 4), ﬁnd all α such that α−1σα= τ ; see the example given after Theorem 3.6.

Problem 3.3 List the conjugacy classes of (i) S3, (ii) S4, (iii) S5,(iv)A3,(v)A4, and (vi) A5. Note that (v) and (vi) are not straightforward, some permutation calculations are needed; for a more detailed explanation, see the subsection on the Class Equations in Chapter 5. Secondly, ﬁnd the normal subgroups of (vii) S4 and (viii) A4.

Problem 3.4 Let σ,τ and ν be cycles in Sn. (i) Show that if σ and τ are disjoint and στ = e, then σ = e = τ . (ii) If σ commutes with τ and τ commutes with ν, does it follow that σ commutes with ν? (iii) Suppose σ and τ belong to SX where X ={1,...,n} and both σ and τ move i ∈ X. Prove that if iσr = iτr for all r ≥ 0, then σ = τ . (iv) Let p be a prime. Show that if σ p = ι, the identity permutation, then σ = ι or σ is a product (with possibly only one factor) of p-cycles.

Problem 3.5 Give formulas for the orders of the elements of (i) Sn, and (ii) An, and list them when n = 7.

Problem 3.6 Let p and q be primes where p | q − 1, let σ = (1, 2,...,q), and let 12... q τ = , where 0

Problem 3.7 (i) Show that Sn isomorphic to a subgroup of An+2. (ii) Using Theorem 3.11 and its extension given in the example on page 77,show that if J is a simple subgroup of the symmetric group Sn, then it is also a (simple) subgroup of the alternating group An.

Problem 3.8 (Semi-regularity) A permutation σ on X ={1,...,n} is called semi- regular if (a) every element of X is moved by σ , and (b) σ can be expressed as a product of disjoint cycles all of the same length. The identity permutation is also called semi-regular (it is a product of n cycles each of length 1). (A permutation is called regular if it is semi-regular and transitive.) (i) Prove that σ is semi-regular if and only if it is a power of an n-cycle.

(Hint, if σ = (a1,...,ar )(b1,...,br ) ···(d1,...,dr ), consider the cycle

θ = (a1,b1,...,d1,a2,b2,...,d2,a3,...,dr ).) 3.5 Problems 61

(ii) If τ is an n-cycle, show that τ s is a product of (n, s) disjoint cycles each of length n/(n, s) (note that (n, s) denotes the GCD of n and s). Deduce that if p is a prime, then each positive power of a p-cycle is either a p-cycle or the identity permutation.

Problem 3.9 (Maximal Subgroups of Sn) (i) Show that Sk ≤ Sn,for1≤ k ≤ n, and determine a lower bound on the number of copies of Sk that occur in Sn; see also Problem 4.20(ii). (ii) For 1 ≤ k ≤ n,letSk × Sn−k denote the set of elements in Sn with the form of a permutation using the symbols in {1, 2,...,k}=Y only, multiplied by a permutation using the symbols in {k + 1,k+ 2,...,n}=Z only. (This is the direct product of Sk and Sn−k, see Chapter 7—you need to check that Sk Sk × Sn−k, et cetera) Show that

Sk × Sn−k forms a subgroup of Sn.

If k = n − k, this subgroup is maximal in Sn. This can be proved as follows: Show that the subgroup generated by the elements of Sk,Sn−k and a 2-cycle (u, v) where u ∈ Y and v ∈ Z is the whole of Sn; you are not asked to prove this here, but you could try it, say when n = 5 or 6. Note that the same general argument applies if we replace the ﬁrst set Y by an arbitrary k-element subset of {1, 2,...,n}=N provided the second set Z is replaced by its complement in N. (iii) By (ii) we have Sn × Sn is a subgroup of S2n; show that it is not maximal 2 by constructing a new subgroup of S2n of order 2(n!) . (Hint. Begin with Sn × Sn, add an element of order 2, and then show that the new set forms a proper subgroup of S2n.) As in (ii) the new subgroup constructed here is maximal in S2n. The properties described in (ii) and (iii) form part of the O’Nan–Scott Theorem which provides a complete description of the maximal subgroups of the symmetric groups. A number of these subgroups involve so-called wreath products which are introduced on page 156. For further details, the reader should consult Cameron (1999), page 107.

Problem 3.10 (Alternating Group A4) There are four main ways to represent A4 as follows: (a) The group of even permutations on the set {1, 2, 3, 4}, (b) The group with presentation a,b | a2 = b3 = (ab)3 = e ,

(c) The symmetry group of a regular tetrahedron (with four equilateral triangular sides), and (d) SL2(3) ‘factored by its centre’; see Chapters 4, 8 and 12. (i) Using direct calculation show that (a), (b) and (c) deﬁne isomorphic groups, see Problem 4.4(iv) for (d). (ii) Find the subgroups of A4, and indicate which are normal. 62 3 Group Construction and Representation

Problem 3.11 (i) In this problem you are asked to construct a subgroup J of S5 with order 20. Let σ = (1, 2, 3, 4, 5). Choose a 4-cycle τ so that the subgroup J = σ,τ has this order. One method is as follows. Show that the group H = a,b | a5 = b4 = e, ba2 = ab has order 20, then ﬁnd an isomorphism between H and J .It can be shown that J is a maximal subgroup of S5, it is usually called metacyclic and denoted by F5,4, see Theorem 6.18 and Web Section 6.5. Can a similar construction be undertaken in S7? (ii) Using Lagrange’s Theorem (Theorem 2.27) and Problem 2.20 ﬁnd the subgroups of A5 of order less than 10. It does also have proper subgroups of order 10 and 12, but none larger; see page 101.

Problem 3.12 (Permutation Matrices) Permutation theory can be developed as a part of matrix algebra as follows. Given a permutation σ ∈ Sn,then × n permutation matrix Pn is formed from the n × n identity matrix In by permuting its rows by σ , that is, the ﬁrst row (1, 0,...,0) becomes the 1σ th row, the second row (0, 1, 0,...,0) becomes the 2σ th row, and so on. Prove the following properties:

(i) Each row and each column of Pn contains a single ‘1’ and n − 1 zeros. (ii) The determinant of Pn, det Pn, equals ±1. (iii) The inverse of a permutation matrix is a permutation matrix, and so the set of all n × n permutation matrices forms a subgroup of GLn(F ) for all fields F .Is this subgroup normal? (iv) A permutation σ is even if and only if the determinant of the corresponding permutation matrix is positive. Note that if the definition of the determinant function Δ uses the notions of even and odd permutations to determine the signs in the basic sums, then (iv) cannot be applied to define these permutations, but it can be so applied if Δ is defined as a multilinear alternating function which takes the value 1 on the identity matrix; see, for example, Rose (2002), Chapter 4. πi/3 η 0 Problem 3.13 Let η = e and C = − . 0 η 1 (i) Calculate C3,C6 and C−1. (ii) Choose a 2 × 2matrixD to satisfy CD = DC−1 and D2 = C3. (iii) Show that the group generated by C and D subject to the conditions in (ii) has order12, and gives a representation of the dicyclic group Q3.

This construction can easily be extended to give representations of the groups Q2n+1 for all positive integers n.

Problem 3.14 Show that GL2(4) is isomorphic to a subgroup of GL4(2) as follows. First, note that the set of non-zero elements of a ﬁeld of four elements forms a multiplicative cyclic group of order 3. Second, use this fact to show that GL1(4) is isomorphic to a subgroup of GL2(2). Now repeat this procedure in the case under consideration using block multiplication of matrices. A fair knowledge of linear algebra is needed to complete this problem. 3.5 Problems 63

Problem 3.15 Suppose F is a ﬁeld. Let ITn(F ) be the subset of UTn(F ) (the group of n × n upper triangular matrices, see page 54) of those matrices with 1 at each main diagonal entry—they are sometimes called ‘unipotent’, and let IZTn,r (F ) be the subset of ITn(F ) of those matrices whose ith superdiagonals consist entirely of zeros for i = 1,...,r. We shall return to this example in Chapter 10 when discussing nilpotent groups. Show that

(i) ITn(F ) UTn(F ). (ii) IZTn,r (F ) ITn(F ), and IZTn,r (F ) IZTn,r−1(F ) if r = 1,...,n− 1. (iii) If o(F) = p (and so we work modulo p) show that ITn(F ) is a Sylow r r p-subgroup of GLn(F ), that is o(ITn(F )) = p where p is the largest power of p dividing o(GLn(F )); see Section 6.2. (iv) If A ∈ IZTn,r (F ) and B ∈ ITn(F ), then [A,B]∈IZTn,r+1(F ).

Problem 3.16 Use Problem 3.3(vi) and Theorem 2.29(ii) to give another proof of the simplicity of A5. You should begin by noting that the neutral element e belongs to all subgroups.

Problem 3.17 Show that the group Q does not have a ﬁnite generating set.

4 3 2 Problem 3.18 Prove that S4 has the presentation a,b | a = b = (ab) = e . Show also that if the powers 4, 3 and 2 are permuted, then further presentations of S4 are given.

Problem 3.19 (i) Show that the group SL2(p) possesses only one involution when p is an odd prime. Also note Problem 12.5. (ii) Using a suitable computer program (or working by hand) show that the conjugacy classes (Deﬁnition 2.11(iii)) of the group SL2(5) have orders (and number of classes) 1(2), 12(4), 20(2) and 30, with nine classes in all. (iii) Use (ii) to show that the only proper non-neutral normal subgroup of SL2(5) is isomorphic to C2. (iv) The maximal subgroups of SL2(5) are isomorphic to SL2(3), Q5 and Q3 with orders 24, 20 and 12, respectively. Find subgroups in SL2(5) isomorphic to these groups. The group SL2(5) has a special connection with so-called Frobenius complements, this will be discussed in Web Section 14.3.

Problem 3.20 Let D = c,d | c2 = d2 = e , A = c , and K = cd . (i) Show that A ≤ D, K D, A ∩ K = e , and AK = D if o(D) > 2. (ii) Use this to give new presentations of the dihedral groups Dn. Proposition (i) shows that D is isomorphic to a semi-direct product of K by A; see Section 7.3. As noted earlier (page 26) involutions (that is, elements of order 2) play an important role in the theory, especially in simple group theory. This also applies to groups of the form D generated by two involutions; see, for example, Aschbacher (1986), page 242. Note also that the jump from two to three generators 64 3 Group Construction and Representation can have dramatic effect, for very large and complex groups exist which have a generating set with just three involutions—for example, L3(3) with order 5616, see pages 264 and 265, and Problem 12.10. Problems 2.28, 3.21 and 3.23 should also be consulted.

Problem 3.21 (A Presentation of Sn+1 for n>1) Let a1,...,an be symbols satisfying the following conditions, where Qj,k does not apply if n = 2,

P : 2 = ≤ ≤ i ai e for 1 i n, 2 Qj,k : (aj ak) = e for 1 ≤ k

(i) Show that (a) aj and ak commute if 1 ≤ k

Let G be the group generated by the set {a1,...,an} with the relations (∗), let H be the subgroup generated by the subset {a1,...,an−1}, and let Z be the set of cosets

Z ={H,Han,Hanan−1,...,Hanan−1 ...a1}.

(ii) Using (i), show that if Hy ∈ Z and 1 ≤ i ≤ n, then Hyai ∈ Z. There are a number of cases to consider. (iii) Use (ii) to show that [G : H]≤n + 1, and so by induction deduce o(G) ≤ (n + 1)!. (iv) By considering maps from G to Sn+1 of the form ai → (i, i + 1), show that G is a presentation of Sn+1 using Problem 3.1 and (iii).

When Sn+1 is represented in this way, that is, when it is generated by a set of involutions, it is known as a Coxeter Group. The theory of these groups has developed considerably in the past 30 years; see, for example, Suzuki (1982) and Björner and Brenti (2005).

Problem 3.22 Suppose m is a positive even integer.

(i) Let Rm be the subgroup of GL2(C) generated by 0 ω 01 A = and B = , m ω 0 −10

where ω is a primitive 2mth root of unity. Prove Rm Qm; see page 59. (ii) Prove that Qm has a unique involution C, and Z(Qm) = C . (iii) Further, show that Qm/Z(Qm) Dm. (Note that factor groups are discussed in Chapter 4.) If G is a 2-group (one whose order is a power of 2, see Chapter 6) and it has a single subgroup of order 2, then it has been shown that G is either cyclic or dicyclic; see Kurzweil and Stellmacher (2004), page 114. 3.5 Problems 65

Problem 3.23 Working in S8, investigate the subgroup F generated by the permutations a = (3, 4)(5, 6), b = (1, 3)(2, 4)(5, 7)(6, 8) and c = (1, 5)(2, 6)(3, 8)(4, 7). Construct a list of the elements with their orders, and so calculate o(F). Secondly, list the subgroups of F , and determine their orders, their normality status, and which of them is the centre. Finally, ﬁnd a presentation for F , that is, ﬁnd a collection of properties of a,b and c from which all remaining properties can be derived. The calculations can be done by hand, or by using a computer algebra package which includes the basic permutation operations. See also Problem 8.12.

Problem 3.24 (Project—The Group GL2(3)) A number of groups play a special role in the theory, A5 being one of these. Another is the general linear group GL2(3), later (in Chapter 12) we shall describe some of its properties. Here you are asked to give representations in terms of the three main construction methods discussed in this chapter. In particular, (a) ﬁnd three 2 × 2 matrices with orders 2, 3 and 8, respectively, which generate the group, (b) ﬁnd three permutations of the set {1, 2, 3, 4, 5, 6, 7, 8} which generate the group as a subgroup of S8, and (c) show that the group has a presentation in the form a,b,c | a8 = b2 = c3 = e,bab = a3,bcb= c2,c2a2c = ab,c2abc = aba2 .

You will need to ﬁnd inclusion maps between these systems, and then show that each contains 48 elements (Theorem 3.15). Note that there is no unique answer to this problem, and that it can be done by hand, but a suitable computer algebra package would help with the calculations. This project will be continued in Problem 6.23. Chapter 4 Homomorphisms

During the past half century and more, one of the underlying ‘themes’ in mathematics has been the realisation that maps are equally as important as objects or sets, particularly those that preserve or transfer properties from one object or system to another. They have been called natural maps, morphisms, or sometimes structure- preserving maps. We use the first of these names as a general term for these maps. Therefore, as we have introduced groups, our basic objects of study, we must now discuss the natural maps between them. In group theory, they are called homomorphisms, and isomorphisms when the correspondence is ‘exact’; isomorphisms were introduced informally on page 17 in Section 2.1. Two groups can appear to be distinct, but are, in fact, identical from the group- theoretical point of view. For example, consider D3, the dihedral group of the triangle defined on page 3, and S3, the group of all permutations on the set {1, 2, 3}. They have identical group-theoretic properties, and a bijection θ between them that satisfies

ghθ = gθhθ (4.1) articulates this fact; equation (4.1) is called the homomorphism equation.Themap θ : D3 → S3 can be deﬁned as follows (it is not the only possible map; see Sec- tion 4.4). Referring to the deﬁnitions given on page 3, the rotation α ∈ D3 about the centre of the triangle by angle 2π/3 clockwise is mapped to (1, 2, 3), that is, we set αθ = (1, 2, 3).Using(4.1)wehave

α2θ = αθαθ = (1, 2, 3)2 = (1, 3, 2), and

ιθ = α3θ = (αφ)3 = (1, 2, 3)3 = e where ι denotes the ‘identity map’, see page 281. Also, the reﬂection β ∈ D3 about a vertical axis is mapped to (2, 3), that is, we set βθ = (2, 3). Then by (4.1)again

H.E. Rose, A Course on Finite Groups, 67 Universitext, DOI 10.1007/978-1-84882-889-6_4, © Springer-Verlag London Limited 2009 68 4 Homomorphisms we have αβθ = αθβθ = (1, 2, 3)(2, 3) = (1, 3), and α2βθ = α2θβθ = (1, 3, 2)(2, 3) = (1, 2).

It is now easily seen that θ is a bijection and (4.1) holds for all elements of D3;we say that D3 and S3 are isomorphic. Isomorphisms are bijections; on the other hand, there is much to be gained by considering maps that preserve the group operation (they satisfy (4.1)) but are not necessarily bijective, these maps are called homomorphisms. For example, suppose ∗ G1 = GL2(Q), G2 = Q and φ : G1 → G2 is deﬁned by

Aφ = det A for A ∈ G1.

As det A = 0 and det AB = det A det B, for all A,B ∈ G1,themapφ is a homomorphism, that is, it satisfies (4.1). We shall refer to this example repeatedly in the next few sections, and so we call it the standard example. In this chapter, we define homomorphisms, isomorphisms and factor groups, derive their basic properties—the Isomorphism and Correspondence Theorems, discuss the cyclic groups, and introduce automorphisms (that is, isomorphisms of a group to itself). There are two Web Sections, 4.6 and 4.7, the first discusses a special kind of homomorphism called the transfer used in Web Section 6.5, and the second extends our work on presentations introduced in Chapter 3.

Maps—Left or Right

In algebraic contexts, we write maps and functions on the right, and in most cases we do not use brackets; that is, we write

aφ and not φ(a) forthevalueofthemapφ at the argument a. In the western world, we read and write from left to right, and so this is a more natural notation—when applying a map φ to an argument a,weﬁrst choose a in the domain, then we apply φ to obtain the value aφ in the range (co-domain); see Appendix A. This notational convention may seem strange at ﬁrst, but it does make many constructions clearer, especially those involving composition or permutations. Also we use lower case Greek letters for maps or functions (including permutations) throughout. The argument of a map φ is given by the Roman letter or letters immediately to its left. For example, in the expression abcφ the argument is abc, but in the expression aφbψ the argument of ψ is b (and this expression is the product of aφ and bψ). On some occasions we use brackets to aid clarity, so we might write (abc)φ for abcφ,or(aφ)(bψ) for aφbψ,but(aφb)ψ when the argument of ψ is aφb, that is when the argument is the product of aφ and b. 4.1 Homomorphisms and Isomorphisms 69

4.1 Homomorphisms and Isomorphisms

The basic notions are given by

Deﬁnition 4.1 Let G1 and G2 be groups, and let φ be a map from G1 to G2.The map φ is called a homomorphism from G1 to G2 if

ghφ = gφhφ for all g,h ∈ G1.

Note that the product gh on the left-hand side of this equation is in G1, and the product gφhφ (or (gφ)(hφ)) on the right-hand side is in G2.

Deﬁnition 4.2 Let φ be a homomorphism mapping G1 to G2.

(i) φ is called the trivial homomorphism if aφ = e, for all a ∈ G1; (ii) φ is called an isomorphism if it is also a bijection from G1 to G2, the groups 1 G1 and G2 aresaidtobeisomorphic, and we write G1 G2 in this case; (iii) φ is called an endomorphism if it is a homomorphism of G1 to itself; (iv) φ is called an automorphism if it is an isomorphism of G1 to itself.

Isomorphisms were ﬁrst introduced by Jordan in 1865 whilst he was working on his proof of the Jordan–Hölder Theorem; see Chapter 9. We use the word ‘trivial’ in (i) above to imply that the map trivialises, or destroys, all properties of the group except those associated with the neutral element. The identity map ι : G → G which is given by gι = g for all g ∈ G is an example of an automorphism of G. The reader should also refer to the note on ‘isomorphism classes’ on page 17.

Examples One homomorphism was discussed on page 68 (the standard example), four more are given now; see also the examples on pages 17 and 84, and in Prob- lem 4.1.

(a) If G1 = R, G2 is the group of complex numbers having absolute value 1 with the operation complex multiplication, and φ1 : G1 → G2 where

xφ1 = cos x + i sin x,

for x ∈ R, then φ1 is a homomorphism mapping G1 to G2. It is not an isomorphism because xφ1 = (x + 2kπ)φ1 for each k ∈ Z. (b) If G1 = Z/6Z (the set {0, 1, 2, 3, 4, 5} with operation addition modulo 6), G2 = (Z/7Z)∗ (the set {1, 2, 3, 4, 5, 6} with operation multiplication modulo 7), and φ2 : G1 → G2 is given by a aφ2 = 3 modulo 7,

for a ∈ G1, then φ2 is an isomorphism; the reader should check this.

1Some authors use the symbol =∼ in place of . 70 4 Homomorphisms

∗ (c) If G1 = C , the multiplication group of the non-zero compex numbers, and φ3 : G1 → G1 satisﬁes 2 zφ3 = z ,

then φ3 is an endomorphism. Reader, why is φ3 not an automorphism? (d) If G1 = Z and φ4 : G1 → G1 satisﬁes aφ4 =−a for a ∈ G1, then φ4 is an automorphism of G1.

The basic properties of these maps are given by the following lemmas.

Lemma 4.3 Suppose φ is a homomorphism between the groups G1 and G2 with neutral elements e1 and e2, respectively.

(i) e1φ = e2. −1 −1 (ii) If g ∈ G1 then g φ = (gφ) .

Note that in (ii) the inverse on the left-hand side is in G1, and the inverse on the right-hand side is in G2.

Proof (i) As g = ge1 for all g ∈ G1,wehavegφ = ge1φ = gφe1φ and (i) follows by cancellation. (ii) Using the group axioms and (i) we have −1 −1 e2 = e1φ = gg φ = gφg φ,

the result follows as inverses are unique and two-sided (Theorem 2.5).

From now on we use the symbol e for the neutral element of every group.

Lemma 4.4 Suppose φ : G1 → G2, ψ : G2 → G3, and both φ and ψ are homomorphisms.

(i) φ ◦ ψ is a homomorphism mapping G1 to G3. (ii) The image of φ is a subgroup of G2.

(iii) If H ≤ G1 and φ is the map φ with its domain restricted to H , then φ is a homomorphism from H into G2. (iv) If φ is an isomorphism, so is φ−1.

The image of φ, see (ii), is denoted by im φ.Alsoφ in (iii) is sometimes written φ|H and described as ‘φ restricted to H ’.

Proof Straightforward, see Problem 4.3.

We now deﬁne the kernel, an important entity in the theory. It is a normal subgroup, the reader should review the material of these subgroups given in Chapter 2. 4.1 Homomorphisms and Isomorphisms 71

Deﬁnition 4.5 Let φ be a homomorphism mapping G1 to G2. The subset of G1 deﬁned by

{a ∈ G1 : aφ = e} is called the kernel of φ, and it is denoted by ker φ.

∗ In the standard example φ : GL2(Q) → Q where Aφ = det A (page 68), ker φ is the set of 2 × 2 matrices A with determinant 1, the neutral element of Q∗. Hence ker φ equals SL2(Q) in this example. (Note SL2(Q) GL2(Q).) The basic properties of the kernel are given by

Lemma 4.6 Suppose φ : G1 → G2 is a homomorphism.

(i) ker φ G1, see Deﬁnition 4.5. (ii) φ is injective if and only if ker φ = e .

This lemma is useful in its own right, but it is also useful because it provides a second method (apart from Theorem 2.29) for showing that a subset of a group is a normal subgroup, that is, by showing the subset in question is the kernel of a homomorphism; see, for example, the proof of the N/C-theorem (Theorem 5.26).

Proof (i) The property ker φ ≤ G1 follows from Theorem 2.13, Deﬁni- tion 4.1, and Lemma 4.3. Now by Lemma 4.3 again, if a ∈ G and g ∈ ker φ,

a−1gaφ = a−1φgφaφ = (aφ)−1eaφ = e,

which shows that a−1ga ∈ ker φ. Normality follows by Theorem 2.29. (ii) Suppose ﬁrst φ is injective. If ker φ = e , we can ﬁnd c ∈ G1 such that c = e and cφ = e = eφ; but as φ is injective, this implies c = e which contradicts our assumption. Hence ker φ = e . Conversely, if ker φ = e and bφ = cφ,forb,c ∈ G1, then

− − − e = (cφ) 1bφ = c 1bφ, and so c 1b ∈ ker φ.

But ker φ = e , and so b = c. This holds for all b,c ∈ G1, and so the result follows.

As an example we give a proof of Cayley’s Theorem. It was first proved in 1850, and was a development of some work undertaken by Cauchy during the previous decade; see page 42. Note that there is nothing unique about the symmetric group here, for it can be shown that every group of order n is isomorphic to a group of m × m matrices, for some m ≤ n, defined over an arbitrarily given field, see Prob- lem 4.17. Also, in Web Section 4.7 we show that every group is a factor group of a free group. 72 4 Homomorphisms

Theorem 4.7 (Cayley’s Theorem) Every group is isomorphic to a subgroup of a symmetric group.

Proof Let G be a group. For ﬁxed g ∈ G, deﬁne the map σg : G → G by

aσg = ag for all a ∈ G.

Using cancellation (Theorem 2.6), we see immediately that σg is a bijection; that is, σg is a permutation of the underlying set of G. Also, if g,h ∈ G,

aσgh = a(gh) = (ag)h = (aσg)σh = a(σg ◦ σh) for all a ∈ G. (4.2)

Hence we can deﬁne a map θ : G → SG, where SG is the group of permutations of the elements of the set G,by

gθ = σg for all g ∈ G,

and it is a homomorphism by (4.2). The result follows by Lemma 4.4(ii).

This result is usually not the best possible. For example, the group D4 has order 8 and so, by Cayley’s Theorem, it is isomorphic to a subgroup of S8 (with order 40320). But, in fact, it is isomorphic to a subgroup of S4 (with order 24), see Chap- ter 8. On the other hand, for a few groups the theorem is best possible, for example, Q2 (which also has order 8) is isomorphic to no subgroup of Sn if n<8.

Factor Groups

Cosets were deﬁned in Chapter 2.Hereweask: Is it possible to make the set of cosets of a subgroup H of G into a new group? The answer is yes, but only when H is a normal subgroup of G (Deﬁnition 2.28). Consider the following simple example which is typical of the general situation. Let G = Z and H = 2Z, the even integers under addition. Clearly H G, and there are just two cosets: the even integers 2Z, and the odd integers 2Z + 1. Now

an even integer plus an even integer is even, an even integer plus an odd integer is odd, and (∗) an odd integer plus an odd integer is even.

This suggests that we can treat the set of cosets in this example as a new two element group with the operation given by (∗), and this will characterise the terms ‘even’ and ‘odd’ when applied to the integers. Hence we make the following 4.1 Homomorphisms and Isomorphisms 73

Deﬁnition 4.8 Given K G, g,h ∈ G and using the operation in G, we deﬁne the coset product,orcoset multiplication,ofgK and hK by

gKhK = ghK.

There is an important point concerning this definition which can cause some misunderstanding at first. By Lemma 2.22,ifj ∈ gK then jK = gK; that is, the representative g in the coset gK is not unique. Therefore, in the definition above we must check (in Theorem 4.9 below) that it is a product of cosets considered as entities in their own right, and it does not depend on the coset representatives g and h—we say the product is “well-defined”.

Theorem 4.9 Suppose K G. (i) If aK = a K and bK = b K, then abK = a b K. (ii) The set of cosets of K in G with coset multiplication given by Deﬁnition 4.8 forms a group.

Proof (i) The hypotheses and Lemma 2.22 give k1,k2 ∈ K to satisfy

a = ak1 and b = bk2,

which shows that

a b = ak1bk2.

By Theorem 2.29 and as K is a normal subgroup, we can ﬁnd k3 ∈ K to satisfy k1b = bk3,so

a b = ak1bk2 = abk3k2 ∈ abK, and (i) follows by Lemma 2.22 again. (ii) By (i), the set of cosets is closed under (well-deﬁned) coset multiplication. Using the corresponding properties of G, coset multiplication is associative, the neutral element is K (as eK = K), and the inverse of the coset aK is a−1K (as aKa−1K = aa−1K = K); the result follows.

We give an example to show that normality is essential in this result. Let G = 3 2 2 D3 = c,d | c = d = e, dc = c d and let J = d .WehavecJ ={c,cd} and Jc ={c,c2d}, that is, J is not normal in G. Property (i) also fails, for if we let a = c, a = cd and b = b = c, then aJ = cJ ={c,cd}=a J (as d2 = e), and bJ = cJ = b J .ButabJ = c2J ={c2,c2d} whilst a b J = cdcJ ={d,e}=J which shows that abJ = a b J .

Deﬁnition 4.10 The group of cosets given by Theorem 4.9(ii) is called the factor group, or sometimes the quotient group,ofG by K, and it is denoted by G/K. 74 4 Homomorphisms

In some contexts, G/K is referred to as “G over K”. We shall show below, especially in Theorem 4.17, that this ‘fractional’ notation is a reasonable choice but it needs to be treated with care. However, we can see already by Lagrange’s Theorem (Theorem 2.27) that, if o(G) < ∞, then

o(G/K) = o(G)/o(K).

Example The notation Z/nZ used in Chapter 2 can now be explained. As nZ Z (the groups are Abelian) we can form the factor group Z/nZ using Deﬁnitions 4.8 and 4.10. By the First Isomorphism Theorem to be proved below, this group is isomorphic to the group N of integers modulo n with addition modulo n as its operation, see page 18. The factor group Z/nZ contains n cosets, they are r + nZ, for r = 0,...,n− 1, and the coset product mirrors the standard addition modulo n in N exactly. As Z/nZ is isomorphic to N, we use this (distinctive) notation for both—a slight ‘abuse of the notation’, but as the groups are isomorphic, no problems should arise.

4.2 Isomorphism Theorems

We come now to what is probably the single most important collection of results in the theory—the Isomorphism Theorems. Note that the naming and numbering of the theorems given below is not universally accepted. Although many of the ideas, theorems and proofs had been ‘known’ for some time previously, they were first sys- tematically formulated and proved in detail2 by the German mathematician Emmy Noether (1882–1935) during the 1920s; and one of their first appearances in print was in Moderne Algebra by B.L. van der Waerden. This two-volume work was first published in the 1930s and it has had a considerable impact on the development of algebra in general during the twentieth century. We begin by returning to the example concerning even and odd integers discussed on page 72. Define a map φ : Z → T1 (page 12)by

aφ = 1ifa is even, and aφ =−1ifa is odd.

Clearly, this is a surjective homomorphism mapping Z to T1 with kernel 2Z, and

Z/2Z T1, see (∗) on page 72. This is typical of the general situation given by the following major result.

2A proof of the First Theorem appears in Burnside’s classic text written a quarter of a century before; see also the comment below Deﬁnition 4.2. 4.2 Isomorphism Theorems 75

Theorem 4.11 (First Isomorphism Theorem) If G1 and G2 are groups, and φ : G1 → G2 is a surjective homomorphism with kernel ker φ = K, then

G1/K G2.

Proof By Theorem 4.9 and Lemma 4.6, G1/K is a group. We deﬁne a map θ : G1/K → G2 as follows. If a ∈ G1 then aK ∈ G1/K, and we set

(aK)θ = aφ.

First, we need to show that θ is well-defined. Suppose aK = a K, then by Lemma 2.22 we have a a−1 ∈ K = ker φ, and so a a−1φ = e which gives a φ = aφ. This shows that aKθ = aφ = a φ = a Kθ as required. The map θ is surjective because φ is surjective. It is also injective, for if aKθ = bKθ, then by definition aφ = bφ which gives in turn e = (aφ)−1bφ = a−1bφ and a−1b ∈ ker φ = K. By Lemma 2.22 again, this shows that aK = bK,soθ is injective, and therefore it is a bijection. For the homomorphism property we have (aK)θ (bK)θ = aφbφ by definition = abφ as φ is a homomorphism = (abK)θ by definition = (aK)(bK) θ by coset product,

and the theorem follows.

All is not lost if the homomorphism is not surjective, for we have the corollary given below where the symbol stands for ‘is isomorphic to a subgroup of’.

Corollary 4.12 If ψ : G1 → G2 is a homomorphism with kernel ker ψ = K, then

G1/K G2.

Proof By Theorem 4.11, G1/K im ψ, and by Lemma 4.4(ii) we have im ψ ≤ G2, the corollary follows.

If in this corollary the group G1 is simple, then it is isomorphic to a subgroup of G2 provided ψ is not the trivial homomorphism.

∗ ∗ Examples Let G1 = C , G2 = R , see page 18, and let φ be the absolute value function given by ∗ zφ =|z| for z ∈ C . This is a non-surjective homomorphism with image R+.Nowkerφ is the subgroup of C∗ of those complex numbers which have absolute value 1. Corollary 4.12 gives 76 4 Homomorphisms

∗ ∗ C / ker φ R ; that is, associated with every non-zero complex number there is a unique non-zero real number, its absolute value. This value belongs to R+ a subgroup of R∗. Returning to the standard example we note that the det function is a surjective ho- ∗ momorphism mapping GLn(Q) onto Q with kernel SLn(Q), hence Theorem 4.11 gives ∗ GLn(Q)/SLn(Q) Q . (4.3) Also if we replace the field Q by a finite field F with o(F) = q, then a similar isomorphism applies and we obtain o(GLn(F )) = o(SLn(F ))(q − 1) by Lagrange’s Theorem (Theorem 2.27)asF has q − 1 non-zero elements.

We single out for special mention the following homomorphism.

Deﬁnition 4.13 Let K G, and let φ be the surjective homomorphism G → G/K given by gφ = gK for all g ∈ G, where K = ker φ.Themapφ is called the natural homomorphism, from G to the factor group G/K = im φ.

There are three further isomorphism theorems, see also Section 9.1. Some will not be required until later but as they are all consequences of the First Theorem (Theorem 4.11) we shall present them now. The Second and Third Theorems give conditions under which factor groups can be simpliﬁed, that is, parts cancelled out, whilst the remaining result, called the Correspondence Theorem, is not a single theorem but a collection of results which relate the properties of G ‘above a normal subgroup K’ to the properties of G/K, it or one of its extensions will be used many times in the sequel. We begin with a lemma about intersections and we restate the basic facts concerning products, see Theorem 2.30.

Lemma 4.14 Suppose H ≤ G and K G. (i) H ∩ K H . (ii) HK ≤ G and HK = KH. (iii) HK G if we also have H G. (iv) If J H then JK HK.

Proof (i) By Theorem 2.15,wehaveH ∩ K ≤ H . For normality we argue as follows. Let h ∈ H .Ifj ∈ H then h−1jh∈ H .AlsoasK G, h−1jh∈ K if j ∈ K. Hence if j ∈ H ∩ K, then h−1jh∈ H ∩ K and the result follows by Theorem 2.29. For the remaining proofs, see Theorem 2.30 for (ii) and (iii), and Prob- lem 4.6(iv) for (iv). 4.2 Isomorphism Theorems 77

We come now to the Second Theorem, as it will be used mainly in our work on series in Chapters 9 to 11, it can be omitted on a ﬁrst reading of the early chapters.

Theorem 4.15 (Second Isomorphism Theorem) If H ≤ G and K G, then K HK ≤ G and H/H ∩ K HK/K.

Proof As K G, by Lemma 4.14 we have HK ≤ G. Also clearly K ≤ HK, and so Problem 2.14(ii) gives K HK and we can form the factor group HK/K. We deﬁne a map θ : H → HK/K by

hθ = hK for h ∈ H.

As hkK = hK for all h ∈ H and k ∈ K, it follows that θ is a surjective map. It is also a homomorphism for if h1,h2 ∈ H we have using coset multiplication (Deﬁnition 4.8)

h1h2θ = h1h2K = h1Kh2K = h1θh2θ.

Hence the conditions for the First Isomorphism Theorem (Theorem 4.11) apply, and so to prove the theorem we need to show that ker θ = H ∩ K.Now h ∈ ker θ if, and only if, hK = K. If this equation holds then h ∈ K, and so h ∈ H ∩ K. Conversely, if h ∈ H ∩ K, then h ∈ K and hK = K clearly follows. Therefore, Theorem 4.11 now gives

H/ker θ = H/H ∩ K HK/K.

Note that one consequence of this result is: If H ≤ G, K G and H ∩ K = e , then HK/K H .

Example We give a proof of a permutation result to illustrate the use of this theorem, see page 49. We show that if G ≤ Sn and G contains an odd permutation, then exactly half of the elements of G are even and half are odd. By Theorem 3.11, An Sn, and G ≤ Sn by hypothesis, so the Second Isomorphism Theorem gives

G/(G ∩ An) GAn/An = Sn/An.

This last equation follows because GAn contains all even and at least one odd permutation, and so has an order larger than o(Sn)/2, which gives GAn = Sn by Prob- lem 2.19.Now,aso(Sn/An) = 2 (Theorem 3.11), we have

o(G ∩ An) = o(G)/2; that is exactly half of the elements of G belong to An.SoifG is simple and G

Correspondence Theorem

The Correspondence Theorem which some authors call the ‘Third Isomorphism Theorem’ will be considered now. It is an important result with many applications, and it is a direct consequence of the First Isomorphism Theorem (Theorem 4.11). As noted above, it has a number of parts, some of which will be added later. Suppose θ is a surjective homomorphism mapping G1 to G2 with ker θ = K; that is, θ maps G1 onto G2, and K onto e . Informally, the theorem says that the properties and structure of that part of G1 which lies above K is exactly mirrored by the properties and structure of G2, and vice versa. This is illustrated in the diagram below.

θ G1 −→ G2 ·· θ H −→ Hθ and G1/H G2/H θ if H G1 ·· −1 θ −1 Jθ −→ J and G2/J G1/J θ if J G2 ·· θ K −→ e

There are also some upward inclusions. If H satisﬁes K ≤ H ≤ G1, then Hθ ≤ G2, −1 and if H G1, then Hθ G1;alsoifJ ≤ G2, then K ≤ Jθ ≤ G1, et cetera. Hence we have

Theorem 4.16 (Correspondence Theorem) Suppose G1 and G2 are groups, H ≤ G1, J ≤ G2, and θ is a surjective homomorphism mapping G1 to G2 with ker θ = K. This implies G1θ G2 and Kθ = e .

(i) If K ≤ H ≤ G1, then Hθ ≤ G2. (ii) If K ≤ H G1, then Hθ G2 and G1/H G2/H θ. −1 (iii) K ≤ Jθ ≤ G1. −1 −1 (iv) If J G2, then Jθ G1 and G1/J θ G2/J .

Proof The proof is mainly a matter of checking that group axioms or subgroup conditions hold, we shall establish parts (i) and (iv) and leave the remaining parts for the reader to complete (Problem 4.13). (i) By Lemma 4.3, Hθ is a non-empty subset of G2. Also, if h1,h2 ∈ −1 = −1 ∈ H , then (h1θ) h2θ h1 h2θ Hθ as H is a subgroup of G1 and θ is a homomorphism. This gives (i). −1 (iv) Suppose g ∈ G1 and k ∈ Jθ , and so gθ ∈ G2 and kθ ∈ J .AsJ G2, these properties show that

(gθ)−1kθgθ ∈ J, 4.2 Isomorphism Theorems 79

but (gθ)−1kθgθ = (g−1kg)θ, hence g−1kg ∈ Jθ−1. By (iii), this gives −1 Jθ G1. For the second part, deﬁne the map ψ : G1 → G2/J by

gψ = (gθ)J for g ∈ G1.

Now ψ is surjective as θ is surjective, and it is a homomorphism because

g1g2ψ = (g1g2θ)J = (g1θ)(g2θ)J = (g1θJ )(g2θJ)= g1ψg2ψ,

for g1,g2 ∈ G1. The kernel of ψ is {g ∈ G1 : (gθ)J = J }. By Lemma 2.22, − (gθ)J = J if and only if gθ ∈ J if and only if g ∈ Jθ 1.

−1 As this holds for all g ∈ G1, we obtain ker ψ = Jθ . The result follows by applying the First Isomorphism Theorem (Theorem 4.11)toψ.

Example Suppose G1 = Z (and so all subgroups are normal as this group is Abelian) and G2 = Z/30Z. Secondly, suppose φ represents congruence reduc- tion modulo 30, its kernel K equals 30Z. Now if, for example, H = 10Z, then K H G1, Hθ Z/10Z and both G1/H and G2/H θ are isomorphic to the cyclic group of order 3. As an exercise the reader should take J Z/15Z and apply parts (iii) and (iv) of Theorem 4.16.

The last isomorphism theorem follows easily from (iv) in the result above. It provides a further justiﬁcation for the factor group notation.

Theorem 4.17 (Third Isomorphism Theorem) If K G and K H G, then G/H (G/K)/(H/K).

This is sometimes called the Freshman’s Theorem; see Scott (1964).

Proof In (iv) of Theorem 4.16, put G2 = G/K, J = H/K, and let θ be the natural homomorphism (Deﬁnition 4.13) from G to G/K. Note that H/K G/K, the reader should check this. We also have

− (H/K)θ 1 = H. (4.4)

For if g ∈ (H/K)θ −1, then gθ = hK for some h ∈ H . But by deﬁnition gθ = gK, and so hK = gK, and g ∈ hK ⊆ H by Lemma 2.22. This gives (H/K)θ −1 ⊆ H and, as the reverse inclusion is given by deﬁnition, (4.4) follows. Finally, as the conditions in Theorem 4.16(iv) now apply, the result follows.

Referring to the example above, if Cm denotes the cyclic group of order m,we have G1/K C30,G1/H C3 and H/K C10, and Theorem 4.17 shows that C3 C30/C10 (that is 3 = 30/10 !). 80 4 Homomorphisms

4.3 Cyclic Groups

In this section, we discuss the cyclic groups first introduced in Definition 2.19.The elements of a cyclic group are the powers (positive and negative) of a single generator a.IfG is infinite, then all powers are distinct and G is the free group on the set {a},butifo(G) = n and a ∈ G, then an = e, see Theorem 4.20 below. The first theorem provides the basic facts.

Theorem 4.18 (i) For each positive integer n there exists a cyclic group of order n. (ii) All cyclic groups of order n are isomorphic. (iii) All inﬁnite cyclic groups are isomorphic to the group Z. (iv) Every homomorphic image of a cyclic group is cyclic.

Proof (i) We give two examples. The group of integers modulo n, Z/nZ,see page 74, is cyclic of order n. Also the group of the complex numbers e2kπi/n for 0 ≤ k

at φ = bt for t ∈ Z,

then φ is clearly an isomorphism between G and H . (iii) This is similar to (ii). (iv) If G = a and θ : G → H is a surjective homomorphism, then H = aθ , that is H is cyclic with generator aθ.

n We write Cn for the abstract cyclic group of order n, that is, Cn a | a = e , a group with a single generator a, say, and a single relation an = e; see the introduction to Chapter 3.AlsoweuseZ as a notation for an inﬁnite cyclic group, see (iii) above.

Theorem 4.19 (i) The subgroups of Z are e and nZ, one for each positive integer n. (ii) All non-neutral subgroups of an inﬁnite cyclic group are isomorphic to Z.

Proof Suppose H contains a non-zero integer, a,b ∈ H , and H ≤ Z. Then ma + nb ∈ H for all m, n ∈ Z. We can choose m and n so that the greatest common divisor c,say,ofa and b satisﬁes c = ma + nb, c | a, c | b and c>0, hence c ∈ H ; see Appendix B. (We usually use the notation (a, b) for c.) This shows that the least positive integer d,say,inH is a divisor of every element of H , and so H = dZ, an inﬁnite cyclic group. Both parts of the result follow by Theorem 4.18(iii).

Theorem 4.20 Suppose G is a cyclic group of order n. It contains a cyclic subgroup H of order m if and only if m divides n, and when this happens H is unique. 4.4 Automorphism Groups 81

This and the previous result show that all subgroups of a cyclic group are cyclic. Also the proof below is not the shortest or most elegant for this result but it is given as an illustration of the Isomorphism Theorems ‘in action’.

Proof We use the First Isomorphism and Correspondence Theorems. Deﬁne amapψ : Z → Cn as follows. Let j be a generator of the group Cn, and for let aψ = j c where a ≡ c(mod n) and 0 ≤ c

We shall see later that the properties of cyclic groups given in Theorems 4.19 and 4.20 are special. In general, a group of order n can have many subgroups of order m, where m | n, or none at all. If m = pr ,n= ms and p s, then the subgroup Cm is called the (unique) ‘Sylow p-subgroup’ of Cn; see Section 6.2.Alsomore generally, if m | n and (m, n/m) = 1, then Cm is called a ‘Hall’ subgroup of Cn.

4.4 Automorphism Groups

Let G be a group and let Aut G denote the set of all automorphisms of G,see Deﬁnition 4.2. This set can be given a group structure using composition as the operation. The composition of two bijections is a bijection (Appendix A) and so the operation is well-deﬁned and closed by Lemma 4.4(i). Composition is associative (Appendix A), the identity map ι on G acts as the neutral element, and the inverse of an isomorphism is also an isomorphism (Lemma 4.4(iv)). Thus Aut G forms a group under composition.

Deﬁnition 4.21 For a group G, the set of automorphisms of G with the operation of composition forms a new group called the automorphism group of G which is denoted by Aut G.

We give an example now and another extended one at the end of the section, we shall also discuss the automorphism groups of the ﬁnite cyclic groups.

Example If G Z, then Aut G C2 because there are only two automorphisms: For all g ∈ G, the ﬁrst maps g → g, and the second maps g → −g; see Prob- lem 4.19. 82 4 Homomorphisms

Some automorphisms are given by conjugation (Deﬁnition 2.28). Let h ∈ G be ﬁxed, then the map τh : G → G given by

−1 gτh = h gh, for g ∈ G, is an automorphism of G, the reader should check this. An automorphism of this type is called inner, and the set of all inner automorphisms of G is denoted by Inn G. It is a subgroup of Aut G as the following lemma shows.

Lemma 4.22 (i) If G is a group, then Aut G is also a group. (ii) Inn G Aut G.

Proof (i) This was proved above. (ii) The identity map is an inner automorphism (put h = e in the deﬁnition), so Inn G is not empty. For g,h ∈ G,letτg and τh be the corresponding inner automorphisms, and let a ∈ G. Then

−1 −1 a(τgh) = (gh) a(gh) = h (aτg)h = (aτg)τh = a(τg ◦ τh).

As this holds for all a ∈ G, we see that τg ◦ τh = τgh ∈ Inn(G).Also −1 = ≤ ∈ (τg) τg−1 (reader, check). Hence Inn G Aut G. Normality. Let a,g G, θ ∈ Aut G, and aθ−1 = b, that is, bθ = a. By Lemma 4.3(ii), −1 −1 −1 a θ τgθ = aθ τg θ = (bτg)θ = g bg θ −1 −1 = g θbθgθ = (gθ) a(gθ) = aτgθ.

−1 This holds for all a ∈ G, hence θ τgθ = τgθ ∈ Inn G, the result follows.

Using Lemma 4.22, we can form the factor group Aut G/ Inn G;itiscalledthe outer automorphism group denoted by Out G. Following normal practice, an automorphism which is not inner is called outer, so an outer automorphism is an element of one of the cosets of Aut G/ Inn G except Inn G itself. Note that if G is Abelian then Inn G = e , and so all automorphisms except the identity automorphism are outer. Also Corollary 5.27 gives a formula for Inn G. It was conjectured by O. Schreier (1901–1929) that Out G is soluble (for a definition see Section 11.1) if G is simple. This has been shown to be true for finite groups but only by using CFSG (which is discussed in Chapter 12). The conjecture is false for non- 3 simple groups, for example, Out((C2) ) GL3(2), a non-Abelian simple group; see page 264. For a particular group, it can be major task to find its automorphism group. Some examples are given in Chapter 8. For example, consider Aut Sn. It has been shown that, if n = 2 or 6, then Aut Sn Sn and all automorphisms are inner (Problem 8.5). This is not true for S6 which does possess outer automorphisms, and the order of Aut S6 is twice the order of S6; see Problem 4.20. An account of these results is given in Rotman (1994), pages 156 to 162. 4.4 Automorphism Groups 83

Next in this section we show how to construct the automorphism group of a cyclic group. In some cases, it is also cyclic but not always.

∗ Theorem 4.23 Aut Cn (Z/nZ) .

Proof Let g be a generator of Cn. A homomorphism mapping Cn to itself will map g to some power of g, so we deﬁne, for m ∈ Z,

m θm : Cn → Cn by gθm = g .

r r rm Note that as θm is a homomorphism g θm = (gθm) = g for all integers r. As gn = e we only need to consider m in the range 0 ≤ m

r rm 1−sn g θm = g = g = g,

n rt t as g = e. This shows that g θm = g for t = 1,...,n, hence θm is surjective. u um Conversely, if θm is surjective, then g = g θm = g ,forsomeu ∈{1,...,n}. This implies that g1−um = e and so 1 − um is a multiple of n. This can only happen if (m, n) = 1. Therefore, θm is an automorphism if and only ∗ if (m, n) = 1. Hence the map θm → m for m ∈ (Z/nZ) gives the required isomorphism.

The groups (Z/nZ)∗ are Abelian, in Chapter 7 we shall show that they can be expressed as ‘direct products’ of cyclic groups. Also using the theory of primitive roots, a topic from number theory discussed brieﬂy in Appendix B,wehave The group (Z/nZ)∗ is cyclic if and only if n = 2, 4,ps ,or2ps where p is an odd prime and s is a positive integer, and their orders are as follows: o((Z/2Z)∗) = 1, o((Z/4Z)∗) = 2, and o((Z/psZ)∗) = o((Z/2psZ)∗) = ps−1(p − 1).

So in particular, the automorphism group of Cps where p is an odd prime and s is a positive integer is itself cyclic and has order ps−1(p − 1); a result due to Gauss. Further automorphism results are given in Problem 4.18, for elementary Abelian groups, and in Problems 6.18 and 7.10. Overleaf we end this section with an extended example which illustrates some of the methods used to construct automorphism groups, it also provides some more examples of isomorphisms. 84 4 Homomorphisms

Example Show that Aut D4 D4 (also see Problem 4.19). 4 2 3 Let the dihedral group D4 be given by a,b | a = b = e, bab = a .Notefirst that an automorphism maps an element of order k to an element of order k (this follows from Problem 4.5(i) and Lemma 4.4(iv)). Also D4 has exactly two elements of order 4, that is a and a3, and so automorphisms either map a → a and a3 → a3, or map a → a3 and a3 → a. In either case, a2 → a2, and so a2 is fixed by all automorphisms. Secondly, note that b,ab,a2b and a3b all have order two and so an automorphism could map b to b,ortoab,ortoa2b,ortoa3b. This suggests defining φ and ψ by φ : a → a and b → ab, ψ : a → a3 and b → b. These give + akφ = ak,akbφ = ak 1b, and akblψ = a3kbl, where k is to be read modulo 4, and l modulo 2. It is now an easy exercise to check that both φ and ψ are automorphisms, the reader should do this. The remaining automorphisms can be generated using φ and ψ. Clearly, ψ2 = ι, the identity map (automorphism). Also bφ2 = (bφ)φ = abφ = aφbφ = a2b, and similarly bφ3 = a3b and bφ4 = b, and so φ4 = ψ2 = ι. For the remaining condition, we have

aψφψ = (a3φ)ψ = a3ψ = (aψ)3 = a9 = a = aφ3, bψφψ = (bφ)ψ = abψ = aψbψ = a3b = bφ3.

This shows that ψφψ = φ3. No other automorphisms are possible, see the comments in the paragraph above, and so the result follows.

Unlike both S3 and S4 (Problems 4.19(i) and 8.5), D4 has both inner and outer automorphisms, see Corollary 5.27.Asa−1aa = a and a−1ba = a2b, we see that the automorphism φ2 corresponds to conjugation by a, and as bab = a3 and b3 = b, the automorphism ψ corresponds to conjugation by b.Butφ has no such correspondence, and so forms an outer automorphism. By Lemma 4.22(ii), the set of inner automorphisms forms a normal subgroup of Aut D4 which is isomorphic to 2 2 2 2 2 the 4-group T2 (page 19) in this case because (φ ) = ψ = ι and φ ψ = ψφ .The four outer automorphisms are φ,φ3,φψ and φ3ψ.

4.5 Problems

Problem 4.1 Show that the following maps are homomorphisms. (i) The maps (a), (b), (c), and (d) given on pages 69 and 70. (ii) The trivial map (Deﬁnition 4.2). (iii) The sgn map from Sn to C2 (Deﬁnition 3.7 on page 47). 4.5 Problems 85

∗ (iv) The determinant map from GL2(F ) to F where F is a ﬁeld (page 68). (v) The projection map from R2 to R given by (x, y) → x.

Problem 4.2 Show that the following pairs of groups are isomorphic.

(i) GL2(2) and S3, see Problem 2.20. (ii) Let F be a field. The first group is F ∗, the multiplicative group of F , and the second group has underlying set F1 = F \{1} and the operation ∗ where a ∗ b = a + b − ab for a,b ∈ F1. First, you will need to show that (F1, ∗) is a group. (iii) Q+ and the additive group of all polynomials in the variable x with integer coefficients. (Hint. Use Unique Factorisation (Theorem B.6).)

Problem 4.3 (i) to (iv) Give proofs of the four parts of Lemma 4.4. (v) Suppose G is a group and X is a set. Given a bijection θ : G → X, construct an operation on X so that θ becomes an isomorphism of G to the group formed by X with this operation. Show also that this operation is unique.

Problem 4.4 An exercise working with cosets. Let G = SL2(3). (i) Find Z(G) and show that it has order 2. (ii) Write out a list of representatives of the cosets of Z(G) in G. (iii) Show that if E is a coset of Z(G) in G and E = Z(G), then either E2 = Z(G) or E3 = Z(G), where Ek is deﬁned using the coset product. Count the number of solutions in each case. (iv) Use (iii) and Problem 3.10 to show that G/Z(G) A4, that is, SL2(3) can be treated as an extension (Deﬁnition 9.9)ofC2 by A4.

A similar argument can be used to show that the general linear group GL2(3) is an extension of C2 by S4.

Problem 4.5 (i) Suppose φ : G → G satisﬁes gφ = g−1 for all g ∈ G. Show that φ is a homomorphism if and only if G is Abelian. (ii) Let G be a ﬁnite group and let θ : G → G be an automorphism. Further suppose (a) if g ∈ G and gθ = g, then g = e, and (b) θ 2 is the identity map ι on G. Use these to show that gθ = g−1 for all g ∈ G, and so deduce that G is Abelian. (Hint. First show that {a−1 · aθ : a ∈ G}=G.)

Problem 4.6 (Abelian Factor Groups and the Derived Subgroup) Before tackling this problem the reader should revisit Problem 2.16 which gives the basic properties of the derived subgroup.

(i) Show that every factor group of an Abelian group is Abelian. (ii) Prove that if K G, then

G/K is Abelian if and only if G ≤ K, 86 4 Homomorphisms

where G denotes the derived (or commutator) subgroup of G.Thisisanim- portant fact we use many times. = (iii) Use (ii) to show that Sn An.

(iv) Suppose H1,H2 ≤ G, H2 H1 and J G. Show that JH2 JH1, and JH1/J H2 is Abelian if H1/H2 is Abelian.

Problem 4.7 Suppose K G and o(G/K) = n<∞. (i) Show that if g ∈ G, then gn ∈ K. (ii) Suppose (m, n) = 1, g ∈ G and gm ∈ K, prove that g ∈ K.

Problem 4.8 (Perfect Groups) A group G is called perfect if it equals its derived subgroup, that is, if G = G . Show that (i) An equivalent deﬁnition is: No non-neutral factor group is Abelian. (ii) If H ≤ G and H is perfect, then H ≤ G .

Problem 4.9 Let θ : G → H be a homomorphism. Prove the following: (i) If a ∈ G, then (an)θ = (aθ)n, for all n ∈ Z. (ii) If K G, K ⊆ ker θ and, for a ∈ G, θ : G/K → H is deﬁned by

(aK)θ = aθ,

then θ is a homomorphism. You need to show that θ is well-deﬁned. (iii) If j1,j2 ∈ G, then [j1,j2]θ =[j1θ,j2θ].

Problem 4.10 Suppose G is a ﬁnite group with the property

(ab)n = anbn, for all a,b ∈ G where n is some ﬁxed integer larger than 1. (i) Let n n n Gn = a ∈ G : a = e and G = c : c ∈ G . n Using a suitably chosen homomorphism show that both Gn and G are normal n subgroups of G, and deduce o(G ) =[G : Gn]. (ii) Show that (a) if n = 2 then G is Abelian, and (b) if n = 3 and 3 o(G), then G is again Abelian. More is known, see Alperin (1969).

Problem 4.11 Let G be a ﬁnite group with a normal subgroup K satisfying o(K),[G : K] = 1.

Using the Isomorphism Theorems show that K is the unique subgroup of G with order o(K) by considering what happens to another such subgroup K1 in G/K;see 4.5 Problems 87

Theorem 6.10.Ifπ denotes the set of prime factors of o(K), then K is called the π-radical of G, see Section 10.2, it is usually denoted by Oπ (G).

Problem 4.12 Suppose K1,...,Kn are normal subgroups of G.LetL = G/K1 ×···×G/Kn, the direct product of G/K1,...,G/Kn; see Section 7.1.(We can treat L as the group of ordered n-tuples {g1K1,...,gnKn} for gi ∈ G with component-wise multiplication.) Deﬁne a map θ : G → L by

gθ = (gK1,...,gKn) for g ∈ G. n Show that θ is a homomorphism, and deduce G/ i=1 Ki is isomorphic to a subgroup of L.

Problem 4.13 (i) Give proofs of the second and third parts of the Correspondence Theorem 4.16. (ii) Suppose K G. Prove the following extension of the Correspondence The- orem: G/K is simple if and only if K is a maximal normal subgroup of G; that is, K is a proper normal subgroup of G, and no other normal subgroup J of G exists which satisﬁes K

Problem 4.14 (Coset Enlargement) (i) Suppose J and K are normal subgroups of G, and J ≤ K.Letξ : G/J → G/K be deﬁned by (aJ )ξ = aK for a ∈ G. Show that ξ is a well-deﬁned surjective homomorphism with kernel K/J.Themapξ is called the enlargement of cosets map. (ii) Use (i) to reprove the Third Isomorphism Theorem (Theorem 4.17).

Problem 4.15 Let G be a ﬁnite Abelian group. (i) If p | o(G), show how to ﬁnd an element g ∈ G of order p.(Hint.Write o(G) = pn and use induction on n.) A second proof of this result is given in Theorem 6.2. (ii) If o(G) = mn, (m, n) = 1, H,J ≤ G, o(H) = m, o(J) = n and both H and J are cyclic, show that G is also cyclic. (iii) Give an example to show that the statement in (ii) is false if G is not Abelian.

Problem 4.16 (Properties of the Centre) Derive the following properties of the centre Z(G) of a group G. Note that by Problem 2.14(ii) a subgroup of Z(G) is normal in G. Suppose K G throughout. (i) Construct an example to illustrate the following fact: There exists at least one group G with proper subgroups H and J which have the properties: e

(iii) If o(K) = 2 then K ≤ Z(G). (iv) If θ : G → J is a surjective homomorphism and H ≤ Z(G), then Hθ ≤ Z(J). (v) Show that if H is an Abelian subgroup of G, then HZ(G) is also an Abelian subgroup of G. (vi) Z(K) G; does it follow that Z(K) ≤ Z(G) (see Problem 4.22)? (vii) If K ≤ J ≤ G, then [J,G]≤K if and only if J/K ≤ Z(G/K). (viii) Finally, show that Z(G) G ≤ Z . Z(G) ∩ K K

Problem 4.17 Suppose F is a field and G is a finite group. Show that G is isomorphic to a subgroup of the general linear group GLn(F ),forsomen not larger than o(G), using the following method. Let U denote the collection of all expres- = ∈ sions of the form z g∈G mgg where mg F . Apply component-wise addition and scalar multiplication to show that U forms a vector space over F . Secondly, for h ∈ G define a map θh : U → U by

zθh = mggh. g∈G

Show that θh deﬁnes an invertible linear map on U, and the collection of these maps forms a group isomorphic to GLn(F ) where n is the dimension of the vector space U. This result can of treated as a ‘matrix version’ of Cayley’s Theorem (Theorem 4.7), note that there is no restriction on the choice of ﬁeld F .

Problem 4.18 (Elementary Abelian Groups) An Abelian group all of whose non- neutral elements have order p, for some ﬁxed prime p, is called an elementary Abelian group (or sometimes an elementary Abelian p-group; see Chapter 6).

(i) Let F denote the field Z/pZ (page 18), and let G be an elementary Abelian p-group. On G define an ‘addition’ ⊕ by x ⊕ y = xy,forx,y ∈ G, and a ‘scalar multiplication’ expressed using concatenation by cx = xc,forc ∈ F and x ∈ G. Prove that G with these new operations forms a vector space over F . We denote it by G. (ii) Show that the subgroups of G correspond to the subspaces of G. (iii) If G is finite, then G will have a finite basis and finite dimension m,say.(This can be proved using some basic linear algebra, or it follows from Theorem 6.3.) Show that the automorphisms of G correspond to the linear maps on G; and so deduce that

Aut G GLm(p).

In Problem 7.14, we shall show that G is a direct product of m copies of Z/pZ,this will provide another proof of the result given in (i). 4.5 Problems 89

Problem 4.19 (i) Find the automorphism groups of the following groups: (a) Z, (b) the 4-group T2,(c)S3,(d)C4, and (e) Cpn where p is an odd prime and n is a positive integer. (ii) Is Aut(D8) D8? (iii) Suppose o(G) < ∞. Show that Aut(G) = e if and only if o(G) ≤ 2. (Hint. Show that the map deﬁned by a → a−1 is an automorphism of an Abelian group.)

Problem 4.20 (i) Let the map ψ : S6 → S6 satisfy:

(1, 2)ψ = (1, 5)(2, 3)(4, 6), (1, 3)ψ = (1, 4)(2, 6)(3, 5), (1, 4)ψ = (1, 3)(2, 4)(5, 6), (1, 5)ψ = (1, 2)(3, 6)(4, 5), (1, 6)ψ = (1, 6)(2, 5)(3, 4).

Beginning with Problem 3.1(i), this can be extended to an automorphism of S6; you are not asked to prove this but you could consider what is needed. Using this 2 fact show that ψ equals the identity map on S6, and so provides an example of an outer automorphism of this group. For more details, see Rotman (1994), pages 156 to 167. It can be shown that Aut Sn Sn provided n = 2 or 6. When n = 2 use (iii) in Problem 4.18. But it is a fact that S6 is the only non-Abelian symmetric group which has an outer automorphism; see page 82.

(ii) Use (i) and Problem 3.21 to show that S6 has 12 subgroups isomorphic to S5, note that six of them possess no 2-cycles and act transitively on six points. One member of this second set of six subgroups can be constructed as follows: Let H be the group generated by (1,i + 1)ψ for i = 1, 2, 3, 4, then choose four (of ten) products of three 2-cycles in H which satisfy the conditions of Problem 3.21 with n = 4.

Problem 4.21 The outer automorphism group for the group A5 is isomorphic to C2. Find an outer automorphism for A5, and consider what would be needed to establish the previous statement. (Hint. Use Theorem 3.6.)

Problem 4.22 (Project—Characteristic Subgroups) In this project, you are asked to develop the notion of characteristic subgroup which is similar to, but stronger than, normality. A subgroup H of a group G is said to be characteristic in G if Hφ ≤ H for all φ ∈ Aut G; this is denoted H char G. (Note that Hφ = {hφ : h ∈ H }.) Prove the following statements. (i) It is sufﬁcient to require Hφ= H . (ii) Normality is equivalent to: Hν ≤ H for all inner automorphisms ν : G → G. 90 4 Homomorphisms

(iii) Unlike normality, the characteristic relation is transitive. (iv) If J char K and K G, then J G, but does this proposition also follow if J K and K char G? (v) A subgroup of a cyclic group is characteristic. (vi) Z(G) char G; we say “the centre of a group is a characteristic subgroup of G”. (vii) The derived subgroup G of G is a characteristic subgroup of G. (viii) Give an example of a normal subgroup which is not characteristic. Chapter 5 Action and the Orbit–Stabiliser Theorem

There is only a small intersection (mainly involving examples) between the material in this chapter and the next, with that in Chapter 7. Hence Chapter 7, which contains work on direct products and Abelian groups, can be read first. In the last chapter, we introduced homomorphisms, they are maps that transfer properties from one group to another, and they satisfy the homomorphism equation (4.1). Here, given a set X we introduce new collections of maps that transform X to itself and which are governed by a group G; that is, for each g ∈ G we define a map \g : X → X, and map composition ‘corresponds’ to the group operation. The map \g is a permutation of X and it is called an action of G on X.Inmany cases, X is closely related to G, but not always. It is also possible to define an action using homomorphisms; see Theorem 5.12. In one sense, this important notion has been part of the theory since its inception, but only in particular instances. If you look for the word ‘action’ in Scott’s group theory text published in 1964—the standard introduction to the theory at that time—you will not find it. But you will find many entities now associated with actions, for example, ‘centraliser’ and ‘normaliser’. About this time, and due in part to the work of Wielandt (1964), it was realised that a number of constructions have a similar basis, and emphasising their similarity would give new insights into the theory. Also at around this time, elegant proofs of some major theorems based on new and easily defined actions appeared; for instance, McKay’s proof of Cauchy’s Theorem (Theorem 6.2) or the main proof of the First Sylow Theorem (Theorem 6.7) both given in the next chapter. Nowadays actions form an important part to any introduction to group theory. We begin this chapter by defining actions and two major associated entities: orbits and stabilisers. We then prove the OrbitÐStabiliser Theorem which leads to a number of important applications. Three major examples follow that introduce centralisers and normalisers, vital entities especially for the finite theory. In Web Section 5.4, we extend our work on permutations begun in Chapter 3, discuss transitive and primitive permutations, and prove Iwasawa’s simplicity lemma, this work has applications in Chapter 12.

H.E. Rose, A Course on Finite Groups, 91 Universitext, DOI 10.1007/978-1-84882-889-6_5, © Springer-Verlag London Limited 2009 92 5 Action and the OrbitÐStabiliser Theorem

5.1 Actions

We begin by considering an example. Let G be the group (Z/7Z)∗, and let X = {1, 2, 3, 4, 5, 6}. In this particular example, the set X and the group G have the same elements, and so we have underlined these elements when they are being treated as members of X. Consider the (right) multiplication of an element x of X by an element g of G, that is, x · g. Later we shall write this as x\g.Wehave

x · e = x, that is, the (right) multiplication of elements of X by e does not alter X. Also, by associativity we have x · (gh) = (x · g) · h, where g,h ∈ G, that is, applying gh to x is the same as ﬁrst applying g to x, and then applying h to the result. We call this procedure an action of G on X,see Deﬁnition 5.1 below. Further, as 4 ∈ G we have

1 · 4 = 4, 2 · 4 = 1, 3 · 4 = 5, 4 · 4 = 2, 5 · 4 = 6 and 6 · 4 = 3, that is, the set X has been permuted by this procedure. There is nothing special about ‘4’, the calculation works for all elements of G; the reader should try one. This procedure gives a map from the group G to the set SX of all permutations of X, where X is the underlying set of G; see Theorem 5.2 below. The term G-set is sometimes used for the set X (that is, when G is acting on X), we will not use this notation because in some cases X and G are unrelated. With the example above in mind we begin by stating the basic

Deﬁnition 5.1 Given a non-empty set X and a group G,wesayG acts on X if, for each g ∈ G, there exists a map \g : X → X, and these maps satisfy

(i) x\e = x, (5.1) (ii) x\(gh) = (x\g)\h, for all x ∈ X and g,h ∈ G. We call the map \g an instance of the action of the group G on the set X.

Notes (a) By (5.1), the group operation ‘corresponds’ to composition of actions, that is, the map \gh is defined to equal \g ◦\h. (b) We usually follow the convention that groups and elements of groups are denoted by letters at the beginning of the alphabet, and sets and elements of sets are denoted by letters at the end of the alphabet. (c) More formally, we can rewrite Definition 5.1 as follows: The function \ that maps the set of pairs X × G to X, and which satisfies the two parts of (5.1), is called an action of G on X, see Theorem 5.12. (d) The action defined above is a ‘right action’; we could also define a left action but as we write functions on the right, we shall only consider the former. 5.1 Actions 93

(e) Many authors use either xg, or sometimes xg,forx\g. Both concatenation and the exponential notation are used widely in many contexts, so for the sake of clarity, it seems preferable to use a new symbol, those readers used to the old concatenation notation can simply ignore the backslash. Personally, the author ﬁnds the exponential notation particularly confusing. Also we use the concatenation notation xg for the particular action called the natural action (Example (a) below), and so we need a distinct notation for the general case. In a recently published book by Isaacs (2008), the author uses x • g for our x\g.

We shall give some examples below, but ﬁrst we prove the following basic result.

Theorem 5.2 Using the notation set out above, the map \g : X → X is a permutation (bijection) of the set X.

Proof Suppose x,y ∈ X and x\g = y\g. Then by (5.1)wehave x = x\e = x\ gg−1 = (x\g)\g−1 = (y\g)\g−1 = y\ gg−1 = y,

that is, the map \g is injective. Secondly, suppose z ∈ X then, for g ∈ G, z = z\e = z\ g−1g = z\g−1 \g,

that is, z\g−1 is a preimage of z under the map \g. Hence this map is also surjective, and so it is a permutation of X.

Examples We give four here, three more extended examples will be discussed in the next section; see also the proofs of Cauchy’s and the First Sylow Theorems given in Chapter 6. (a) Let G be a group and let X be the underlying set of G. The group G acts on X by right multiplication if we deﬁne, for g ∈ G and x ∈ X,

x\g = xg,

see the example given at the beginning of this section. This clearly satisfies the conditions (5.1), and the corresponding action is called the natural action on G. (b) Let V be a vector space defined over a field F . The multiplicative group F ∗ of F acts on V by scalar multiplication,forifa,b ∈ F ∗ and v ∈ V , the standard vector space axioms give

v\1 = v1 = v and v\(ab) = v(ab) = (va)b = (v\a)\b.

(c) Let G =e and X be an arbitrary set, then G acts on X if we deﬁne x\e to equal x for all x ∈ X. (d) Let G = Sn and let X ={1,...,n}, then if we deﬁne x\σ = xσ for σ ∈ G and x ∈ X, we obtain a new action called the permutation action; see Theorem 5.2. 94 5 Action and the OrbitÐStabiliser Theorem

There are two important entities associated with an action which govern its basic properties. They are orbits and stabilisers, we introduce orbits ﬁrst. Let the group G act on the set X. Deﬁne a relation ∼ on X by: If x,y ∈ X, then

x ∼ y if and only if x\g = y for some g ∈ G.

To put this informally, x is related to y if we can ‘get from x to y’ by using an element of G.

Lemma 5.3 The relation ∼ deﬁned above is an equivalence relation.

Proof We have x ∼ x as x\e = x. Secondly, suppose x ∼ y, that is, x\g = y for some g ∈ G. Then y\g−1 = (x\g)\g−1 = x\ gg−1 = x\e = x,

and so y ∼ x. Finally, suppose we also have y ∼ z with y\h = z for some h ∈ G. Then x\(gh) = (x\g)\h = y\h = z, that is, x ∼ z.

Deﬁnition 5.4 (i) An equivalence class of the equivalence relation ∼ given in Lemma 5.3 above is called an orbit of the action of G on X. The orbit containing the element x ∈ X is called the orbit of x, and it is denoted by OG{x},orO{x} when it is clear which group G is being used. (ii) An action of G on X is called transitive if there is only one orbit, that is, X itself, otherwise it is called intransitive.

By Lemma 5.3, X is a disjoint union of its orbits (Appendix A). The orbit of x ∈ X, O{x}, is the subset of X of those elements that we can ‘get to’ starting with x and applying elements of G (that is, by applying the maps associated with the elements of G), so an action is transitive if we can ‘get from’ every member of X to every other member of X by applying elements of G. For instance, the action in Example (d) opposite is transitive because Sn contains all 2-cycles—if y,z ∈ X, then the 2-cycle (y, z) belongs to Sn and this 2-cycle maps y to z.But if in this example we change the group to (1, 2) C2, then the action would not be transitive if n>2 because, for instance, no element of this group maps 1 to 3. Also, if we replace X by X1 ={1,...,n,n+ 1}, the action of Sn on X1 would again be intransitive because no permutation in Sn maps 1 to n + 1. An extension of transitivity is as follows. An action of G on X is called k-transitive if for all pairs of k-element subsets Y and Z of X, there exists an element g ∈ G such that \g is a bijection between Y and Z. For example, the action given in Example (d) is k-transitive if k ≤ n. See Web Section 5.4 and Section 12.4. We have deﬁned both orbits and cycles (Deﬁnition 3.3), the notion of an orbit in a general group is related to the notion of a cycle in a symmetric group. If n>1 and 5.1 Actions 95

σ ∈ Sn then, by Theorem 3.4, σ can be expressed as a (disjoint) product of cycles

σ = (a1,...,aj )(b1,...,bk) ··· where j + k +···=n.LetH =σ , the cyclic subgroup of Sn generated by σ , then H acts on X ={1,...,n} by setting x\h = xh for h ∈ H and x ∈ X.Hereh has the t t form σ for some t ∈ Z, and so the orbit of a1,say,is{a1σ : t = 0, 1, 2,...} which is exactly the cycle (in σ ) containing a1. Hence in this particular example, orbits and cycles coincide.

Our second new entity is the stabiliser which we introduce by

Deﬁnition 5.5 GivenagroupG actingonasetX (Deﬁnition 5.1), and x ∈ X,the subset of G, g ∈ G : x\g = x , is called the stabiliser of x in G; it is denoted by stabG(x).

The stabiliser of x is the subset (subgroup, see Lemma 5.6)ofG of those elements whose associated maps ‘do not move’ x. For instance, if we let x = 1 in Example (d) on page 93, then the stabiliser of 1 is the set of all permutations in Sn which leave the element 1 ﬁxed. This is clearly a subgroup of Sn isomorphic to Sn−1, and is an instance of

Lemma 5.6 Using the notation set out in Deﬁnition 5.5, for x ∈ X,

stabG(x) ≤ G.

Proof Clearly, e ∈ stabG(x) as x\e = x by (5.1). If g,h ∈ stabG(x), then x\g = x = x\h and, by (5.1)again,wehave x\ gh−1 = (x\g)\h−1 = (x\h)\h−1 = x.

−1 This shows that gh ∈ stabG(x), now use Theorem 2.13.

We come now to the OrbitÐStabiliser Theorem, the main result in this chapter. It has a number of applications and, considering its importance, it is remarkably easy to prove.

Theorem 5.7 (OrbitÐStabiliser Theorem) If G acts on a set X, x ∈ X, and OG{x} is the orbit of x, then o OG{x} = G : stabG(x) . 96 5 Action and the OrbitÐStabiliser Theorem

Proof Note ﬁrst stabG(x) ≤ G by Lemma 5.6. We deﬁne a map γ from OG{x} to the set of right cosets of stabG(x) in G, and the theorem follows by showing that this map is a bijection. The map γ is given by (x\g)γ = stabG(x) g,

where g ∈ G, and so x\g ∈ OG{x}. First, we show that this map is well- deﬁned. Suppose x\g = x\h, then as in the proof above we have x\ gh−1 = (x\g)\h−1 = (x\h)\h−1 = x\e = x

−1 by (5.1), that is, gh ∈ stabG(x). By Lemma 2.22, this gives (stabG(x))g = (stabG(x))h, as required. Second, note that γ is clearly surjective, for if g ∈ G, then one preimage of (stabG(x))g is x\g. Last, we show that it is injective. Suppose stabG(x) g = stabG(x) h, −1 −1 then, as above, this shows gh ∈ stabG(x), and so x\(gh ) = x. Hence x\h = x\gh−1 \h = x\ gh−1h = x\g,

that is, γ is injective. The theorem now follows.

Referring back to Example (d) on page 93,letx = n. The orbit of n is {1,...,n} O{ } = (as the action is transitive), and so o( n ) n. We noted above that stabSn (n) Sn−1. Hence we have by the OrbitÐStabiliser Theorem, and as o(Sn) = n!, O{ } = =[ : ]= : o n n Sn Sn−1 Sn stabSn (1) .

This is a easy example but it does provide an illustration of the theorem ‘at work’. As a second application of this theorem we prove the following useful result. Another proof was given in Problem 2.27. Note that if H and J are both non-normal subgroups of G, then HJ is not a subgroup of G, see the example below. But if either H or J is normal, then HJ is a subgroup (Theorem 2.30) and the result follows using the Second Isomorphism and Lagrange’s Theorems.

Theorem 5.8 If G is a ﬁnite group and H,J ≤ G, then

o(H J )o(H ∩ J)= o(H)o(J).

Proof We deﬁne an action. Let X ={Hg : g ∈ G}, the set of right cosets of H in G. The subgroup J acts on X by right multiplication if we set

Hg\j = Hgj for j ∈ J.

This is an action since Hg\e = Hge = Hg and Hg\j1j2 = Hgj1j2 = (Hgj1)\j2 = (Hg\j1)\j2. The orbit O{H} of H is {H \j : j ∈ J }, and this 5.1 Actions 97

equals HJ, and so o(HJ) = o(H) × o(O{H}). (Note that HJ is a disjoint (by Lemma 2.23) union of right cosets of H , and the orbit of H under this action is the union of those cosets of H we can ‘get to’ starting with H itself and applying elements j in J .) Further, the stabiliser of H ,stabJ (H ), equals {j ∈ J : Hj = H}, and so stabJ (H ) = H ∩ J (as Hj = H if and only if j ∈ H ). Hence, using the equation for o(HJ) above, the OrbitÐStabiliser Theorem gives

o(HJ) o(J) = o O{H} = J : stab (H ) = , o(H) J o(H ∩ J)

using the equation for stabJ (H ) and Lagrange’s Theorem (Theorem 2.27)for the last identity, the result follows.

3 2 2 Example Suppose G = D3 a,b | a = b = e, a b = ba, H =b≤G and J =ab≤G. Then o(H) = o(J) = 2, H ∩ J =e, and so o(HJ) = 4, by Theo- rem 5.8. Clearly, HJ is not a subgroup of G (as 4 6, also neither H nor J is normal), but it is a union of two cosets. For as ab = ba2,wehaveHab= Hba2 = Ha2 and so HJ = H ∪ Ha2. Note that we also have HJ = J ∪ bJ .

We have shown (Theorem 5.2) that an action of a group G on a set X is a collection of permutations of X; that is, the action provides a map from G to SX.We shall develop this further.

Deﬁnition 5.9 Let the group G act on the set X.Themapν : G → SX given by

gν =\g, for all g ∈ G, is called the permutation representation of G for this action.

Lemma 5.10 The map ν given by Deﬁnition 5.9 is a homomorphism.

Proof For all g,h ∈ G and x ∈ X,wehaveby(5.1),

x\(gh) = (x\g)\h = x(\g ◦\h)

by the deﬁnition of composition (◦) of functions. The lemma follows as this holds for all x ∈ X.

This leads to the following useful

Theorem 5.11 Let G act on X with permutation representation ν as deﬁned above, then ker ν = stabG(x). x∈X 98 5 Action and the OrbitÐStabiliser Theorem

Proof This is an immediate consequence of the deﬁnitions. As ν is a homomorphism (Lemma 5.10), its kernel is the set of those g ∈ G for which \g is \ = ∈ the identity permutation in SX, that is, x g x for all x X. But stabG(x) is ∈ \ = the set of those g G for which x g x, hence x∈X stabG(x) is the set of those g ∈ G for which x\g = x for all x ∈ X; that is, the kernel of ν.

∈{ } Referring again to Example (d) on page 93,ifk 1,...,n , then stabSn (k) is the set of all permutations which fix k. Hence the intersection of these stabilisers for k = 1,...,n,ise, and so the kernel of the permutation representation in this case is the neutral subgroup. This reflects the fact that Sn has very few normal subgroups. The converse of the last result is also valid as we show now. It can be used as an alternative definition of the action of a group G on a set X.

Theorem 5.12 Suppose σ : G → SX is a homomorphism of G to the group of all permutations on the set X. The map deﬁned by \g = gσ, for all g ∈ G, is an action of G on X, and the permutation representation of this action is identical to σ .

Proof For x ∈ X,wehavex(eσ) = x (as eσ is the identity permutation ι on X) and, by composition of maps and as σ is a homomorphism, x(gσ) (hσ ) = x(gσ ◦ hσ ) = x (gh)σ .

Hence, if we deﬁne x\g = x(gσ), for all g ∈ G and x ∈ X, these equations show that \g is an action of G on X.Letν be the permutation representation of this action, that is, for g ∈ G and x ∈ X, x\g = x(gν). Combining these facts gives x(gν) = x(gσ) for all x ∈ X, hence gν = gσ for all g ∈ G, which shows that ν = σ .

Restricted Actions

Here we ask: What is the relationship of stabH (x) to stabG(x) when H ≤ G?To answer this question we ﬁrst need to consider the subset of those elements which are ﬁxed by an action.

Definition 5.13 Let G act on the set X.Weset fix(G, X) = x ∈ X : x\g = x for all g ∈ G ; it is called the fixed set of X under the action of G. 5.2 Three Important Examples 99

Note that fix(G, X) is a subset of X, and so it is not a group; for example, it is empty when the action is transitive, also we have the equivalent definitions fix(G, X) = x ∈ X : O{x}={x} = x ∈ X : stabG(x) = G . (5.2)

If G and X are given by the first example in this chapter (page 92), then fix(G, X) =∅, but if we change X to X ={1,...,7}, then fix(G, X) ={7}. Let G act on a set X and let H ≤ G,wesayH acts on X by restriction of the action of G on X if we ignore those maps \g in the action of G on X where g ∈ H , and only consider those maps \h where h ∈ H . This is clearly an action because H is a (sub)group. For example, if G = Z, H = 2Z, X is the underlying set of G (the integers) and the action of G on X is the natural one given by x\g = xg, then the orbit of x under the action of G is the set of all integer multiples of x, whilst the orbit of x under the restricted action (by H ) is the set of all even integer multiples of x. We have, for x ∈ X and H ≤ G,

stabH (x) = stabG(x) ∩ H, (5.3) by deﬁnition of the restricted action. We also have

Lemma 5.14 If H ≤ G, G acts on X, and H acts on X by restriction of the action of G, then

x ∈ ﬁx(H, X) if and only if H ≤ stabG(x).

Proof We have

x ∈ ﬁx(H, X) if and only if stabH (x) = H by (5.2) if and only if stabG(x) ∩ H = H by (5.3) if and only if H ≤ stabG(x),

by Problem 2.5.

For an example, see Problem 5.3(i). One consequence of this problem is: If G is a p-group (Section 6.1) and p o(X), then there exists x ∈ X whichismovedbyno g ∈ G, that is, ﬁx(G, X) is not empty.

5.2 Three Important Examples

In this section we discuss three action examples, the ﬁrst involves cosets, and the second and third use conjugation of elements and subgroups, respectively. All three introduce major new concepts and theorems which will be used widely in the following chapters, and as noted above all three have a long history in the theory. 100 5 Action and the OrbitÐStabiliser Theorem

Coset Action

For the ﬁrst example, choose H , a subgroup of G, and let X be the set of right cosets of H in G.Giveng ∈ G and Hx ∈ X, we deﬁne

(H x)\g = Hxg. (5.4)

This is an action because (H x)\e = Hxe= Hx and (H x)\g \h = (H xg)\h = Hxgh= (H x)\(gh).

Further, it is a transitive action. For if Hx,Hy ∈ X, then (H x)\x−1y = (H x)x−1 \y = Hy, and so there is only one orbit, that is, X itself. Also

−1 stabG(H x) = x Hx because stabG(H x) = g ∈ G : (H x)\g = Hx = g ∈ G : Hxg= Hx = g ∈ G : xgx−1 ∈ H = x−1Hx, by Problem 2.23. Hence, using the OrbitÐStabiliser Theorem (Theorem 5.7), we obtain −1 G : x Hx = G : stabG(H x) = o(X) =[G : H ] (5.5) because there is only one orbit which in this case is the set of right cosets of H in G, this gives another proof of Problem 2.23(ii). If νH is the permutation representation of this action, then by Theorem 5.11, −1 ker νH = x Hx, x∈G see the comment below Deﬁnition 2.21. The entity on the right-hand side of this equation is called the core of H in G, core(H ) which was deﬁned in Problem 2.24; it is the largest normal subgroup of G contained in H .If[G : H ]=n<∞, then SX Sn, and νH gives a homomorphism from G into Sn. Hence we have

Theorem 5.15 (i) If H

Proof (i) This follows immediately from Theorem 5.11 and Corollary 4.12. (ii) Both of these properties follow from (i) and Problem 2.15(i) with J = core(H ). 5.2 Three Important Examples 101

This is a useful result, especially when n is small, for it shows that if a group G has only a few normal subgroups, then its total number of subgroups is also restricted.

Example Subgroups of A5.IfG is simple and H n!, G does not contain a subgroup (normal or not) of index n. For instance, consider G = A5 with order 60. Suppose H4, the theorem shows that A5 cannot have a subgroup of index 2, 3 or 4, so it cannot contain a subgroup of order 30, 20 or 15. In this case, we say that A5 is not reverse Lagrange. It does contain a number of subgroups of order 12 (with index 5). The group A4 is also not reverse Lagrange, reader why?

We give an application of Theorem 5.15 here, more will follow later.

Theorem 5.16 If G is an infinite group, H ≤ G, and [G : H ] < ∞, then G contains a normal subgroup K which satisfies K ≤ H and G/K is finite.

= −1 Proof Take K g∈G g Hg in Theorem 5.15 and use Problem 2.23.The factor group G/K is ﬁnite because this theorem gives an injective homomorphism into a ﬁnite symmetric group.

This shows that if an infinite group has a subgroup of finite index (and so is infinite), then it also has a normal infinite subgroup of finite index—an important fact concerning infinite groups. This also shows that an infinite simple group has no subgroups of finite index.

Centralisers and Class Equations

Our second extended example involves conjugation, and introduces a number of new concepts and constructions. Let G be a group and let X be the underlying set of G.Ifg,x ∈ G, then g−1xg is called the conjugate of x by g (Definition 2.28). The group G acts on its underlying set G by conjugation if we define − x\g = g 1xg, (5.6) for all g,x ∈ G. The operation defined by (5.6) is an action; for clearly x\e = e−1xe = x, and we have x\(gh) = (gh)−1xgh = h−1 g−1xg h = h−1(x\g)h = (x\g)\h. 102 5 Action and the OrbitÐStabiliser Theorem

The action (5.6) is called the conjugacy action on G, three important entities are associated with it as follows.

Deﬁnition 5.17 Using the conjugacy action deﬁned above, the orbit of x under this action is called the conjugacy class of x and it is denoted by CG{x}, that is, −1 CG{x}= g xg : g ∈ G .

This is, of course, the set of conjugates of x in G. When it is clear which group is involved, we write C{x} for CG{x}.

Deﬁnition 5.18 The stabiliser of x in G under the conjugacy action (5.6) is called the centraliser of x in G, it is denoted by CG(x); that is, −1 CG(x) = stabG(x) = g ∈ G : g xg = x .

Note this is equivalent to CG(x) ={g ∈ G : xg = gx}; therefore, the centraliser of x in G is the subgroup (Lemma 5.6) of those elements g ∈ G which commute with x.

Applying the OrbitÐStabiliser Theorem (Theorem 5.7) we have directly

Theorem 5.19 Using the conjugacy action (5.6) deﬁned above, if x ∈ G, then o CG{x} = G : CG(x) .

By Lagrange’s Theorem (Theorem 2.27), this shows that the order of a conjugacy class of a finite group G divides the order of G. It also shows that the set of elements that commute with a fixed element a, say, forms a subgroup. Both of these facts have important ramifications; for the first, see Lemma 5.21 below. The third entity associated with the conjugacy action is the centre, see Defini- tion 2.32.Ifτ denotes the permutation representation of the conjugacy action defined above, then the kernel, ker τ , is just the centre of the group Z(G); that is, Z(G) = ker τ = CG(x). x∈G

Note also that, using Deﬁnitions 2.32 and 5.17,wehave

x ∈ Z(G) if and only if CG{x}={x}. (5.7)

Example We construct the conjugacy classes, centralisers and centre of the dihedral group D3.LetD3 be given by (Section 3.4)

a,b | a3 = b2 = e,ba = a2b . 5.2 Three Important Examples 103

We have −1 2 C{a}= g ag : g ∈ D3 = a,a ,oC{a} = 2, 2 CD (a) = g ∈ D3 : ga = ag = e,a,a =a,oCD (a) = 3, 3 3 C{b}= b,ab,a2b ,oC{b} = 3, ={ }= = CD3 (b) e,b b ,oCD3 (b) 2.

The reader should check these statements and complete the remaining cases, they provide applications of the OrbitÐStabiliser Theorem. Note ﬁnally that Z(D3) =e ∩ = as CD3 (a) CD3 (b) e . Putting these ideas together we can introduce the Class Equations by

Theorem 5.20 (Class Equations) Suppose G is a ﬁnite group, and CG{y1},..., CG{yk} is a complete list of the conjugacy classes of G whose orders are larger than 1. k ˙ ˙ (i)G= Z(G) ∪ CG{yi} (disjoint unions). i=1 k k (ii)o(G)= o Z(G) + o CG{yi} = o Z(G) + G : CG(yi) . i=1 i=1

Proof (i) This follows immediately from Lemma 5.3 and (5.7) using Deﬁni- tion 5.17—G is a disjoint union of its conjugacy classes. (ii) The ﬁrst equation is given by (i) as the unions are disjoint, and the second follows from Theorem 5.19.

The equations in (i) and (ii) above are called the Class Equations for G.Also the positive integer o(Z(G)) + k, that is, the number of conjugacy classes of G including the singleton classes counted in o(Z(G)), is called the class number of G and it is denoted by h(G). See Appendix C for some examples. Next we give some applications of the Class Equations, more will follow later.

Lemma 5.21 If p is prime and o(G) = pn, then Z(G) = e.

This lemma has important implications for p-group theory; see Section 6.1.

Proof If G is Abelian there is nothing to prove, and if not, then at least one yi exists with o(CG{yi})>1. Referring to the notation given in Theorem 5.20, we have by Theorems 5.19 and 2.27 p G : CG(yi) ,

for i = 1,...,k, as these indices are larger than 1 by deﬁnition of yi . Hence by the second Class Equation we have p | o(Z(G)) because p | o(G) by hypothesis. Now o(Z(G)) > 0(ase ∈ Z(G)), and so this shows that o(Z(G)) ≥ p and therefore o(Z(G)) cannot equal 1; the result follows. 104 5 Action and the OrbitÐStabiliser Theorem

Theorem 5.22 (i) If o(G) = pn, then G is not simple, provided n>1. (ii) If p is prime and o(G) = p2, then G is Abelian.

If o(G) = p, then G is simple and cyclic (and so Abelian) by Theorem 2.34, and (ii) deals with the case o(G) = p2. But there exist non-Abelian groups with order pn for n>2; see Section 6.1 and Problem 6.16.

Proof (i) This follows by Lemmas 2.31 and 5.21 as e

In Chapter 7, we show that if o(G) = p2, then G is cyclic or a ‘product’ of two cyclic groups of order p, that is, elementary Abelian, see Problem 4.18. Centralisers played a vital role in the solution of CFSG, see Chapter 12. Brauer and Fowler (1955) showed that for a given ﬁnite group G there can only be ﬁnitely many simple groups H which contain an involution (element of order 2) a and have the property:

CH (a) G.

In many cases, G can be taken to be quite small and only a few simple groups H are involved, several simple groups can be characterised using this result; see the discussion on page 265. We can extend the notion of centraliser of subsets of a group as follows.

Deﬁnition 5.23 The centraliser CG(X) of a subset X of a group G is given by CG(X) = g ∈ G : ag = ga for all a ∈ X .

The basic properties are as follows; proofs are left as exercises for the reader, see Problem 5.8. Note that as the size of X increases, the size of CG(X) decreases.

(i) If X ⊆ G then CG(X) ≤ G. = (ii) CG(G) Z(G). ≤ = (iii) If H G then CG(H ) h∈H CG(h). (iv) If J ≤ H ≤ G and H ≤ CG(J ), then J ≤ Z(H). If we refer back to the example on page 102 we see that, using (iii) above, = = { } = CD3 ( a ) a , CD3 ( b ) b and CD3 ( a,b ) e as the reader can easily check. See also Problem 5.26. 5.2 Three Important Examples 105

Normaliser

Our last extended action example also uses conjugation but now applied to subgroups. This will introduce the normaliser—one of the most important entities in group theory, and one with a long history in the theory. If H ≤ G and g ∈ G, then g−1Hg = g−1hg : h ∈ H , is a subgroup of G which is called a conjugate subgroup of H in G (Problem 2.23). Using this notion we can deﬁne an action of G on the set of all subgroups H of G by − H\g = g 1Hg for g ∈ G. (5.8) This is an action because H\e = e−1He = H and, for g,h ∈ G, H \gh = (gh)−1Hgh = h−1(g−1Hg)h = (H \g)\h. The orbit of H under this action is the set of subgroups of G which are conjugate to H in G (Problem 2.23 again), and the stabiliser is given by

Deﬁnition 5.24 For H ≤ G, the stabiliser of H in G under the action (5.8) deﬁned above, that is, −1 stabG(H ) = g ∈ G : g Hg = H , is called the normaliser of H in G, and it is denoted by NG(H ).

The normaliser NG(H ) clearly contains H , and it is the largest subgroup of G in which H is normal; the reader should check this, see Problem 5.13. Note that NG(e) = G = NG(G), so in particular it is not true in general that if H

Lemma 5.25 Suppose H ≤ G.

(i) H NG(H ) ≤ G. (ii) NG(H ) = G if and only if H G. (iii) The number of conjugates of H in G equals [G : NG(H )].

−1 Proof (i) Clearly H ≤ NG(H ),ash Hh= H if h ∈ H ;alsoNG(H ) ≤ G by deﬁnition and Lemma 5.6. Theorem 2.29 gives normality. −1 (ii) If NG(H ) = G use (i), and if H G, then g Hg = H for all g ∈ G, that is NG(H ) = G. (iii) This follows immediately from the OrbitÐStabiliser Theorem (Theo- rem 5.7). 106 5 Action and the OrbitÐStabiliser Theorem

Example Let G = S4 and H =(1, 2, 3) C3.Wehave

NG(H ) =(1, 2, 3), (1, 2) S3, for (1, 2)(1, 2, 3)(1, 2) = (1, 3, 2), and so (1, 2) ∈ NG(H ), et cetera. Hence H< NG(H ) < G. Sometimes H equals NG(H ) (in this case, we say H is self- normalising), for example, when H is a maximal non-normal subgroup of G.In other cases, H

The ﬁnal pair of results in this section are easily proved and have a number of useful applications; see Section 4.4 for the basic properties of automorphisms.

Theorem 5.26 (N/C-Theorem) Suppose H ≤ G.

(i) CG(H ) NG(H ). (ii) NG(H )/CG(H ) is isomorphic to a subgroup of Aut H .

The factor group NG(H )/CG(H ) is called the automiser of H in G.

−1 Proof For g ∈ G,letφg be the (inner) automorphism given by aφg = g ag for a ∈ G, and deﬁne θ : NG(H ) → Aut H by

jθ = φj |H for j ∈ NG(H ); (5.9)

that is, jθ is φj with its domain restricted to the subgroup H . Note that jθ ∈ −1 Aut H because j ∈ NG(H ),asj hj ∈ H for h ∈ H , and φ|H is deﬁned by conjugation. Also θ is a homomorphism (as the reader can check). Now

a ∈ ker θ if and only if φa|H is the identity map on H, if and only if h−1ah = a for all h ∈ H, if and only if a ∈ CG(H ),

using Deﬁnition 5.23. Hence ker θ = CG(H ), (i) follows by Lemma 4.6, and (ii) follows by Corollary 4.12.

Example For this example, the reader will need to refer to Section 8.2 where the group SL2(3) is discussed; we show here that this group has a factor group iso- A G = SL ( ) H = 22 , 12 Q morphic to 4; see Problem 3.10.Let 2 3 and 21 22 2. We have H G, and so NG(H ) = G, CG(H ) = Z(G) C2, and Aut H S4 (Problem 6.18). In this case, the N/C-theorem gives G/Z(G) Aut H S4.But o(G/Z(G)) = 12, and we have G/Z(G) A4 because the only subgroup of S4 of order 12 is A4 (Problem 3.3(vii)). 5.3 Problems 107

The last result in this chapter provides an essential ﬁrst step in the construction of the automorphism group of a group.

Corollary 5.27 G/Z(G) Inn G.

Proof Put H = G in Theorem 5.26.WehaveNG(G) = G (as the normaliser of a subgroup always contains that subgroup), CG(G) = Z(G) (see (ii) on page 104), and θ ∈ Inn G using (5.9) with H = G. Note also that the map θ given in (5.9) is surjective in this case.

For example, this corollary shows that Inn S4 S4 as this group is centreless. In fact, we have Aut S4 S4, but this is harder to prove, see Problem 8.5. Some further applications are given in Chapter 8, and in the example at the end of Section 4.4.

5.3 Problems

Problem 5.1 (i) Suppose G ≤ S4 and G acts naturally on the set {1, 2, 3, 4}, see (d) on page 94. Construct the orbits and stabilisers of this action when (i) G =(1, 2, 3), (ii) G =(1, 2, 3, 4), (iii) G =(1, 2)(3, 4), (1, 3)(2, 4),(iv)G = (1, 2), (3, 4), and (v) G = A4. (vi) Find the orbits and stabilisers of the action given in Example (b) on page 93 when dim V = n and n>1. (vii) Let Q[x1,...,xn] denote the set of all polynomials with rational coefﬁcients in the variables x1,...,xn.Forf(x1,...,xn) ∈ Q[x1,...,xn] and σ ∈ Sn, deﬁne

f(x1,...,xn)\σ = f(x1σ ,...,xnσ ).

Show that this deﬁnes an action of Sn on the set Q[x1,...,xn], and describe the orbits and stabilisers. Hence prove that the order of the set of n-variable polynomials of the form f(x1,...,xn)\σ , where f is ﬁxed, is a divisor of n!.

Problem 5.2 Let G be a group with subgroups H and J , and suppose we have the identity ([G : H], [G : J ]) = 1. Show that (i) G = HJ, and (ii) [G : H ∩ J ]= [G : H ][G : J ]. (Hint. Use Theorem 5.8.)

Problem 5.3 Suppose o(G) = pn where p is a prime, and so G is a p-group, see Section 6.1. (i) If G acts on a finite set X using Definition 5.13 show that o fix(G, X) ≡ o(X) (mod p).

(ii) Prove that if e=J and J G, then J ∩ Z(G) = e—a useful result, note that it provides a generalisation of Lemma 5.21. 108 5 Action and the OrbitÐStabiliser Theorem

Problem 5.4 Suppose G acts on a set X. −1 (i) If x,y ∈ X and y = x\g for some g ∈ G, show that stabG(y) = g stabG(x)g, and so deduce the result: o(stabG(x)) = o(stabG(y)). (ii) For g ∈ G, let fix(g, X) denote the subset of X of those x which are fixed by g (so x\g = x). Show that if m is the number of orbits of this action, then m = 1/o(G) o fix(g, X) . g∈G (iii) Using (ii), show that if 1

Problem 5.5 (i) Use Theorem 5.19 to show that o(G) = 2 when G has just two conjugacy classes. (ii) Show that if G is a ﬁnite non-Abelian group, 1 < [G : H ] < 5 and H ≤ G, then G is not simple. (iii) Let G be a ﬁnite group and let h(G) denote its class number, see page 103. Show that the total number of ordered pairs (a, b) which satisfy ab = ba, where a,b ∈ G,ish(G) · o(G).

Problem 5.6 If p0 is the smallest prime dividing o(G), K ≤ G and [G : K]=p0, prove that K G. (Hint. Use Theorem 5.15.)

Problem 5.7 (i) If C is a conjugacy class of a group G, show that the set C−1 = {a−1 : a ∈ C} is another conjugacy class of G, and vice versa. (ii) Construct the conjugacy classes, and their inverses as given by (i), for the groups D4,A4 and SL2(3).

Problem 5.8 (Properties of the Centraliser) (i) to (iv) Prove the centraliser properties listed on page 104. (v) Let H ≤ G and g ∈ G. Show that CH (g) = CG(g) ∩ H , and −1 −1 g CG(H )g = CG gHg .

(vi) Prove CG(H ) ≤ NG(H ) without using Theorem 5.26. (vii) Let H ≤ G. Show that CG(H ) =e if, and only if, Z(J) =e for all J satisfying H ≤ J ≤ G.

Problem 5.9 (i) Find the centralisers of the elements of D6 and S4, and of the subgroups of S4. (ii) Find the normalisers of the subgroups of S4; see Section 8.1 and Lemma 5.25(iii). (iii) Write out the Class Equations (Theorem 5.20)forA5 explicitly; your answer should include a description of all centralisers involved.

Problem 5.10 Let τ = (1,...,m)be an m-cycle in Sn where n ≥ m>1. Show that × = CSn (τ) τ Sn−m, where we assume that S0 e . The direct product notation × is deﬁned in Section 7.1. 5.3 Problems 109

Problem 5.11 (i) Suppose G is a non-Abelian group. Show that if a ∈ G\Z(G), then aZ(G) is an Abelian subgroup of G which properly contains Z(G). (ii) Suppose H,J ≤ G where H is Abelian. The subgroup H is called maximal Abelian if J is not Abelian whenever H

Problem 5.12 Let A and B be subsets of the group G. Show that

(i) if A ⊆ B then CG(B) ≤ CG(A), (ii) A ⊆ CG(CG(A)), (iii) CG(CG(CG(A))) = CG(A).

Problem 5.13 (Properties of the Normaliser) Let H ≤ G. Prove that

(i) NG(H ) is the largest subgroup of G in which H is normal. −1 −1 (ii) NG(g Hg)= g NG(H )g, for all g ∈ G. (iii) If H ≤ J ≤ G, then NJ (H ) = NG(H ) ∩ J . (iv) If K ≤ G, then [H,K]≤H if, and only if, K ≤ NG(H ). (v) Let p beaprime.IfJ ≤ G, o(J) = pr for some r>0 and p |[G : J ],soJ is a non-Sylow p-subgroup of G (Chapter 6). Prove that J

Problem 5.14 Let D denote the subgroup of diagonal matrices in G = GL2(Q). Find NG(D). Do the same calculation for GL3(Q).

Problem 5.15 (i) Suppose H

Problem 5.16 Throughout this problem K G and we say that K is central if K ≤ Z(G).UsetheN/C-theorem (Theorem 5.26) to prove the following propositions. (i) If G is perfect (Problem 4.8) and K is cyclic, then K is central. (ii) If p0 is the smallest prime dividing o(G) and o(K) = p0, then K is central. (iii) If G is infinite and K is finite, then G/CG(K) is finite. Deduce that if the only finite factor group of G has order 1, and o(K) < ∞, then again K is central.

Problem 5.17 (i) Suppose K H ≤ G and J = CG(K). Show that (a) H ≤ NG(K), and (b) if J is self-normalising (that is, J = NG(J )), then K ≤ Z(H). (ii) Let H ≤ G. Prove that CG(H ) = NG(H ) if and only if H ≤ Z(NG(H )). These properties are used in the statement of Burnside’s Normal Complement The- orem (Theorem 6.17). 110 5 Action and the OrbitÐStabiliser Theorem

Problem 5.18 In this problem, you are asked to show that a group G of order 15 is Abelian using the Class Equations; see also Problem 7.21. The method is as follows: Assume the contrary and use Problem 4.16(ii) to show that Z(G) =e, then use the Class Equations (Theorem 5.20) to show that G has exactly one conjugacy class of order 5 consisting of elements of order 3, and then obtain a contradiction.

Problem 5.19 Suppose G = GL2(3) and so o(G) = 48, see Theorem 3.15 and Problem 6.23. (i) Find the centre Z(G) and show that o(Z(G)) = 2. (ii) Let H denote the set of upper triangular matrices in G (that is, H = UT2(3) ab = = which has elements of the form 0 c where a 0 c and we work modulo 3). Show that Z(G) ≤ H ≤ G, and o(H) = 12. (iii) Prove that core(H ) = Z(G) (Problem 2.24). (iv) Use Theorem 5.15 to prove that G/Z(G) S4, see Problem 4.4.

Problem 5.20 Suppose CG(a) G. Use Problem 2.25 to show that a belongs to a normal Abelian subgroup of G. Is the converse false?

Problem 5.21 Suppose o(G) is ﬁnite, and a,b ∈ G. (i) Show that the number of elements g in G satisfying g−1ag = b equals o(CG(a)). (ii) Deduce o(CG(a)) ≥ o(G/G ) where G denotes the derived subgroup of G. (iii) Now suppose K G, [G : K]=p (p a prime), and c ∈ K with the property: CK (c) < CG(c). Show that if b is conjugate to c in G, then b is also conjugate to c in K.

Problem 5.22 Let r = h(G) be the class number of the ﬁnite group G,see page 103. Show that

(i) If p0 is the smallest prime dividing o(G) and rp0 >o(G), then Z(G) = e. (Hint. Use the Class Equations.) (ii) If G is not Abelian, then r>o(Z(G))+ 1. (Hint. Use Problem 4.16(ii).) (iii) If o(G) = p3 and G is not Abelian, then G = Z(G), o(Z(G)) = p and r = p2 + p − 1. (Hint. Apply Problem 4.6 and Lemma 2.31, and show that no conjugacy class has order p2 using Problem 5.21.)

Problem 5.23 (i) Show that elements of the same conjugacy class have conjugate centralisers. (ii) If n1,...,nr is a list of the orders of the centralisers of elements of distinct −1 +···+ −1 = conjugacy classes of G, prove that n1 nr 1. (iii) Deduce there are only ﬁnitely many groups with class number r using the fact that there are only ﬁnitely many ways of writing a positive integer as a sum of reciprocals, see Problem B.4. (iv) Find all groups with class number 3. 5.3 Problems 111

Problem 5.24 Suppose G is ﬁnite and H ≤ G. Show that − o g 1Hg ≤ 1 + o(G) −[G : H ]. g∈G

Use this result to show that if H ≤ G and H contains at least one element of each conjugacy class of G, then H = G.

Problem 5.25 Use the following method ((a) to (d)) to show that if K An, n>4, and K contains a 3-cycle, then K = An. With Theorem 3.14 this provides a new proof of the simplicity of An when n>4. Throughout suppose σ is a 3-cycle in K.

(a) Show that CSn (σ ) > CAn (σ ). = ∩ (b) Secondly, show that CAn (σ ) CSn (σ ) An. [ : ]= (c) Using Theorem 5.8 and (b), deduce CSn (σ ) CAn (σ ) 2. C { }=C { } (d) Lastly, show that Sn σ An σ , and use Theorem 3.6.

Problem 5.26 (Project—Centralisers and Normalisers of Groups of Order 24) Let G1 = S4 and G2 = SL2(3), see Chapters 3 and 8. First, calculate the centralisers of each of the elements of these groups. Second, calculate the centralisers and normalisers of each of the subgroups. Also check that the properties (i) to (iv) given on page 104 apply. Two major theorems in the theory are connected to the relationship between subgroup centralisers and normalisers: the N/C-theorem (Theorem 5.26) and Burnside’s Normal Complement Theorem (Theorem 6.17). Check that these results apply to the groups G1 and G2, and their centralisers and normalisers. Chapter 6 p-Groups and Sylow Theory

There are important connections between number theory and finite group theory, results in one theory have vital applications in the other, and both theories benefit from this interaction. Lagrange’s Theorem shows that a major invariant of a finite group is its order and the prime factorisation of the order. We shall develop this aspect of the theory here; the number-theoretic results we use are discussed in Appendix B. Elements of order two play a central role, they are called involutions. One reason is that an involution is its own inverse; but this is not the only special property. For example, all groups with exponent 2 (a2 = e for all a in the group) are Abelian (Corollary 2.20), and in Problem 2.28 we showed that a non-Abelian simple group is generated by its involutions. But perhaps the most remarkable fact concerning the number two in the theory is the theorem of Feit and Thompson (1963) which states that no non-cyclic group of odd order can be simple. In fact, more is true—a group of odd order must be ‘soluble’, see Chapter 11. First in this chapter, we consider groups whose orders are powers of a particular prime number p—the p-groups. These groups have a number of useful properties, for example, they have normal subgroups for all orders dividing the group order. Also the number of groups of order pn increases exponentially with n, some details are given on page 118. We have noted previously that the converse of Lagrange’s Theorem is not true in general (page 101). But it is true for prime powers, that is, for subgroups which are p-groups. This is the beginning of the Sylow theory which we develop in the second section of the chapter; it is central to finite group theory. We shall give a number of applications. In Section 6.3, we show that if o(G) has up to three not necessarily distinct prime factors, then G is not simple and it can be determined completely; we also prove a remarkable result due to Frattini—the Frattini argument—and introduce ‘nilpotent groups’, which have many properties in common with p-groups including being ‘reverse Lagrange’. We give some more applications in Web Sec- tion 6.5, these will include a proof of Burnside’s Normal Complement Theorem (Theorem 6.17) and a description of those groups all of whose Sylow subgroups are cyclic, see page 130. Throughout this chapter p denotes a prime number.

H.E. Rose, A Course on Finite Groups, 113 Universitext, DOI 10.1007/978-1-84882-889-6_6, © Springer-Verlag London Limited 2009 114 6 p-Groups and Sylow Theory

6.1 Finite p-Groups

In this section, we develop the basic properties of ﬁnite p-groups, and begin with

Deﬁnition 6.1 Let p be a ﬁxed prime. A group G is called a p-group if all of its elements have orders which are powers of p.

The neutral group e is a p-group for all primes p as p0 = 1. We shall give some more examples at the end of this section. In the ﬁnite case, Deﬁnition 6.1 can be replaced by

G is a ﬁnite p-group if and only if o(G) is a power of p.

This follows from Cauchy’s Theorem (Theorem 6.2) for which we give three proofs. The ﬁrst is due to J. McKay and uses a simple action argument, the second uses the Class Equations (Theorem 5.20) but it has the disadvantage that the Abelian case must be treated separately. The result also follows directly from the First Sylow Theorem (Theorem 6.7). Cauchy proved this result for permutation groups in 1845 as part of a series of papers on the properties of permutations; see Section 3.1.Inall probability, the ﬁrst proof for general groups was given by Jordan in the 1870s.

Theorem 6.2 (Cauchy’s Theorem) If G is a ﬁnite group, p is a prime, and p | o(G), then G contains at least one element of order p.

In fact, G contains at least p − 1 elements of order p. In the case p = 2, there may be a unique element of order 2, for instance, in the quaternion (dicyclic) group Q2 (page 119) or the special linear group SL2(3) (page 172). Note that Cauchy’s Theorem applies to all ﬁnite groups.

Proof We begin by introducing a new action. Let X = (g1,...,gp) : gi ∈ G, for i = 1,...,p, and g1 ···gp = e ,

that is, X is the set of all ordered p-tuples of elements of G whose product is the neutral element. We prove the theorem by applying an action of the group Z/pZ to X, and counting orbits. Note ﬁrst

− o(X) = o(G)p 1. (6.1)

= −1 ··· −1 ∈ For arbitrary g1,...,gp−1 in G,ifwetakegp gp−1 g1 , then gp G, the p-tuple (g1,...,gp) belongs to X, and gp is unique by Theorem 2.5. We deﬁne an action of Z/pZ (the integers with addition modulo p)onX as follows. If (g1,...,gp) ∈ X and a ∈ Z/pZ, then

(g1,...,gp)\a = (ga+1,...,gp,g1,...,ga). 6.1 Finite p-Groups 115

The action of a on an element of X cyclically permutes its entries by a places modulo p.ThesetX is closed under this action because, if (g1,...,gp) ∈ X (and so g1 ···gp = e), then by associativity

−1 ga+1 ···gpg1 ···ga = (g1 ···ga) (g1 ···ga)(ga+1 ···gp)(g1 ···ga) = e.

Also it is easily seen that the action axioms (5.1) are satisﬁed. By the OrbitÐStabiliser Theorem (Theorem 5.7), an orbit of this action has order 1 or p (the divisors of o(Z/pZ)). Suppose there are r orbits of order 1, and s orbits of order p. An element of an orbit of order 1 has the form (g,...,g) for some g ∈ G with gp = e (as all cyclic permutations give the same p-tuple). Now r>0 because there is at least one orbit of this type, namely the orbit of the p-tuple (e,...,e). If this was the only orbit of order 1, then r = 1, and by (6.1),

− 1 + sp = o(X) = o(G)p 1,

as X is the disjoint union of its orbits. But this is impossible because p | o(G) by hypothesis, and so p divides the right-hand side of the equation above whilst it clearly does not divide the left-hand side. Therefore, r>1 (in fact, r is a positive multiple of p), and so there must exist at least one g ∈ G satisfying g = e and o(g) = p.

We give a second proof of this theorem using the Class Equation. As noted above the Abelian case must be established ﬁrst, see Problem 4.15.

Second proof—Non-Abelian case We use induction on o(G). Choose a ∈ G such that a/∈ Z(G), it exists because G is not Abelian. This implies that o(CG{a})>1, and so by Theorem 5.19 [G : CG(a)] > 1, that is, CG(a) is a proper subgroup of G.Ifp | o(CG(a)), then the theorem follows by induction because o(CG(a)) < o(G) and so by the inductive hypothesis CG(a) has an element of order p. Hence we may suppose p o(CG(a)) for all a ∈ G\Z(G). But as p | o(G) by hypothesis, this implies p G : CG(a) ,

for all a ∈ G\Z(G) by Lagrange’s Theorem (Theorem 2.27). Applying this to the second Class Equation (Theorem 5.20)givesp | o(Z(G)).ButZ(G) is Abelian, and so by Problem 4.15, Z(G) contains an element of order p, and hence so does G.

We can now show that our two deﬁnitions of a ﬁnite p-group are equivalent.

Theorem 6.3 If G is ﬁnite and p is prime, then G is a p-group if and only if o(G) = pr for some non-negative integer r. 116 6 p-Groups and Sylow Theory

Proof There is nothing to prove if r = 0. Suppose G is a p-group (Deﬁni- tion 6.1)oforderqt n where q is a prime which does not divide n, t>0, and q = p. By Theorem 6.2, G contains an element of order q, which contradicts the hypothesis. Therefore, q = p and n = 1. The converse follows from Lagrange’s Theorem and Deﬁnition 2.19.

Next we show that the property of being a p-group is preserved by taking both subgroups and factor groups, and vice versa. Very few properties are preserved in this way. Abelianness is not one, but it is true for ﬁniteness and, as we shall show later, it is also true for solubility (Chapter 11).

Theorem 6.4 (i) Subgroups and factor groups of p-groups are p-groups. (ii) If K G, and K and G/K are both p-groups, then G is also a p-group.

Proof (i) The first part follows from the definition. For the second part, suppose g ∈ G and o(g) = pr , then using coset multiplication we have r r K = eK = gp K = (gK)p . Hence the order of gK is a divisor of pr , and so the order of every element of G/K is a power of p. r (ii) If g ∈ G, then (gK)p = K for some integer r by the second hy- r pothesis, hence gp ∈ K.ButK is a p-group (the first hypothesis), and so r o(gp ) = ps for some non-negative integer s, so the order of g is a divisor of pr+s . The result follows as this holds for all g ∈ G.

The p-groups have many important properties, for example, they are reverse La- grange. In fact, the following stronger result follows from Lemma 5.21.

Theorem 6.5 If G is a group with order pr , then G is a ﬁnite p-group (by Theo- rem 6.3) and it has subgroups G0,...,Gr satisfying

(a) G0 =e, Gr = G; (b) for i = 1,...,r, Gi G and Gi−1 Gi ; i (c) for i = 0,...,r, o(Gi) = p .

Proof The proof is by induction on r.Ifr = 1 there is nothing to prove, and if r = 2 the result follows from Lemma 5.21 and Theorem 5.22, for then G is Abelian. Hence we may assume that the result holds for all p-groups of order less than pr with r>1. By Lemma 5.21 again, Z(G) = e, and so Z(G) contains an element y which satisﬁes o(y) = pt for some positive t, and if we t−1 put z = yp , then z ∈ Z(G) and o(z) = p. Set

G1 =z,

we have G1 G because G1 ≤ Z(G) G, see Problem 2.14(ii). r−1 Let H = G/G1, by Lagrange’s Theorem (Theorem 2.27), o(H) = p . The inductive hypothesis provides a sequence of subgroups H0,...,Hr−1 which satisﬁes (a), (b) and (c) with r − 1forr, H for G, and Hi−1 for Gi . 6.1 Finite p-Groups 117

Let θ be the natural homomorphism G → H G/G1, then the Correspon- −1 dence Theorem (Theorem 4.16(iii) and (iv)) gives Gi = Hi−1θ which are subgroups of G containing G1,fori = 1,...,r − 1. Applying this theorem again for i = 0,...,r − 1, we have

Gi+1 G(as Hi H) and Gi Gi+1 (as Hi−1 Hi).

Parts (a) and (b) follow, and (c) follows by Lagrange’s Theorem.

The groups Gi in Theorem 6.5 are in many cases not unique. If ri denotes the i number of subgroups of G with order p , both normal and non-normal, then ri ≡ 1 (mod p). We shall prove this fact for ‘Sylow’ subgroups in the next section, and the general result is given in Rotman (1994). There is an extension to Theorem 6.5 which applies when K is a maximal subgroup of G. In this case, K has prime index and is normal in G (cf. Problem 2.19(i)). We derive these properties now.

Theorem 6.6 Suppose G is a ﬁnite p-group, and K is a maximal subgroup. (i) K G. (ii) [G : K]=p.

Proof (i) By induction on r where o(G) = pr , the result clearly holds when r = 1. By Lemma 4.14(ii), K ≤ KZ(G) ≤ G (the second inequality holds as Z(G) G), and so by the maximality of K we have

KZ(G) = G or KZ(G) = K,

both are possible. If KZ(G) = G, there exists g ∈ Z(G)\K,butg ∈ NG(K) (because g ∈ Z(G)); therefore, K is a proper subgroup of NG(K) which implies, by the maximality of K, that NG(K) = G. In turn, this implies that K G by Lemma 5.25(ii). The second possibility is KZ(G) = K, then Z(G) K by Problem 2.14 and Lemma 2.31.AsK is a maximal subgroup of G,wealsohaveK/Z(G) is a maximal subgroup of G/Z(G) by Problem 4.13(ii). By Lemma 5.21 and as G is a p-group, we have o(Z(G)) > 1, so o G/Z(G) < o(G).

Using the inductive hypothesis this gives K/Z(G) G/Z(G), and the Corre- spondence Theorem shows that K G. Hence in both cases K G. (ii) This follows easily from (i) and Theorem 6.5, see Problem 6.1.

Theorems 10.6 and 10.7 on pages 213 and 214 give a second proof of (i). 118 6 p-Groups and Sylow Theory

A large number of p-groups exist with order pr if r is large, many of which differ only slightly from one another. If v(n) denotes the number of (isomorphism classes of) groups of order n, then the following data has been established for powers of 2 (see Besche et al. 2001):

v(4) = 2,v(8) = 5,v(16) = 14,v(32) = 51, v(64) = 267,v(128) = 2328,v(256) = 56092, v(512) = 10494213 and v(1024) = 49487365422; the second of these equations is established below and in Chapter 7. No exact formula is known for v(n) in general, but the following estimate has been given by Higman and Sims:

3 v(pn) = p2n /27 + O n8/3 . On the other hand, if n has a large number of prime factors and is square-free, then v(n) can be quite small. For all integers m, there are inﬁnitely many integers n, with m prime factors, that satisfy v(n) = 1; this is a corollary of Dirichlet’s Theorem on Primes in Arithmetic Progressions; see, for instance, Rose (1999). As examples we have v(15) = 1, v(105) = v(3 · 5 · 7) = 1, and v(5865) = v(3 · 5 · 17 · 23) = 1; see the table in Appendix D.

Non-Abelian Groups of Order 8

The smallest non-Abelian p-groups have order 8 (Problem 2.20), we consider these now. Groups of order p3, p>2, can be treated similarly (Problem 6.5), and Abelian p-groups will be discussed in Chapter 7.LetG be a non-Abelian group of order 8. If G contains an element of order 8, then G is cyclic, also if every non-neutral element has order 2, then G is Abelian (Corollary 2.20), hence we may assume that G contains an element a, say, of order 4 and, by Problem 2.19(i), a G (this also follows from Theorem 6.6). Choose b ∈ G\a, then the elements of G are

e,a,a2,a3,b,ab,a2b, and a3b.

The reader should check that no two of these elements are equal—for example, if a = ab then by cancellation b = e.Nowasb2 ∈ G, it follows that b2 equals one of the eight elements of G listed above. If b2 = ar b, then b = ar but by deﬁni- tion b/∈a, and if b2 = a or a3, then o(b) = 8, and G is cyclic. Hence as we are considering the non-Abelian case, b2 = e or a2; both are possible. Also, as a G we have

b−1ab = as ∈a, for s = 0, 1, 2, or 3. If s = 0 then a = e,ifs = 1 then G is Abelian, and if s = 2 then e = a4 = (b−1ab)2 = b−1a2b, which implies that a2 = e. Hence there is only one possibility: s = 3, that is, b−1ab = a3. This gives ba3 = ab, and so bat = a4−t b for 6.2 Sylow Theory 119 t = 1, 2, 3. These calculations show that there are at most two (isomorphism classes of) non-Abelian groups of order 8:

4 2 −1 3 D4 = a,b | a = b = e,b ab = a ,

4 2 2 −1 3 Q2 = a,b | a = e,b = a ,b ab = a , see pages 58 and 59. The first group, which is isomorphic to the dihedral group of the square, has five elements of order 2. The second is the quaternion group (sometimes called the first dicyclic group), it contains one element of order 1, one element of order 2, and six elements of order 4. As these groups have different numbers of elements of order 2 they are not isomorphic, and so there are exactly two isomorphism classes of non-Abelian groups of order 8. The group Q2 also has one property that it shares with very few others: it is not Abelian and all of its subgroups are normal (groups with this property are called Hamiltonian), see Problem 7.13.

Historical note In the middle of the nineteenth century, the Irish mathematician W.R. Hamilton (1805Ð1865) was looking for fields which extend the complex numbers, he discovered the quaternions which have some remarkable properties but do not (quite) form a field. They can be defined in a similar manner to the complex numbers (that is, as a vector space, now of dimension 4, over R with a vector multiplication) except that the number i is replaced by three entities i, j and k which satisfy i2 = j 2 = k2 =−1 and ij = k =−ji, and a general quaternion is an entity of the form x +iy +zj +tk where x,y,z,t ∈ R. All field axioms are satisfied except one, commutativity of multiplication fails as can be seen above (a system of this type is known as a division algebra). The oc- tonions which are sometimes called Cayley numbers extend this construction to dimension 8, but in this case both commutativity and associativity fail; some brief details are given in Rose (2002), page 176. The ring of integral quaternions has a similar definition to that for the quaternions given above except now x,y,z,t ∈ Z. The quaternion group Q2 forms the group of units (divisors of 1) of this ring when we map a → i and b → j, for then ab → ij where ij = k.

6.2 Sylow Theory

We have seen earlier that the converse of Lagrange’s Theorem is false—if m | o(G) it does not follow that G has a subgroup of order m. For instance, A5 has no subgroup of order 15, see page 101. But there is a partial converse if we restrict m to be a prime power—if pr | o(G), then G does have a subgroup of order pr , and this is the starting point for the Sylow theory. As for Cauchy’s Theorem, we shall give a number of proofs of this important (existence) result. The ﬁrst uses a new action and is due to Wielandt, and the second uses the Class Equations but as with the second 120 6 p-Groups and Sylow Theory proof of Cauchy’s result, it has the disadvantage that the Abelian case must be dealt with separately. The third proof uses some matrix theory. We begin with the main existence theorem ﬁrst proved by the Norwegian mathematician Ludwig Sylow (1832Ð1918) in 1872.

Theorem 6.7 (Sylow Theorem, Part 1) If G is ﬁnite group with order pr m, where p is prime and p m, then G contains a subgroup of order pr .

Proof There is nothing to prove if r = 0, so we may assume that r>0. The following fact concerning binomial coefﬁcients will be used:

pr m The binomial coefﬁcient is not divisible by p. (6.2) pr pr m r (The numerator of pr has the following factors (p in total): − − pr m pr m − 1 ··· p pr 1m − 1 pr m − (p + 1) ··· p2 pr 2m − 1 − − ··· p pr 1m − pr 1 − 1 ··· pr m − pr − 1

and when this expression is divided by the denominator of the binomial coef- ﬁcient, that is (pr )!, all factors of the form pi are removed by cancellation.) We deﬁne a new action. Let X denote the set of all unordered subsets with r no repetitions of the underlying set of G which contain exactly p elements. pr m As there are pr ways of choosing these subsets, we have by (6.2)

p o(X ). (6.3)

The action of G on X is deﬁned by right multiplication: If X1 ∈ X then

X1\g ={xg : x ∈ X1} for g ∈ G.

r Using cancellation we have o(X1\g) = o(X1) = p , and so X1\g ∈ X ,for all g ∈ G. The action axioms (5.1) follow. Let X1, X2,...denote the orbits of this action, as X is a disjoint union of its orbits, we see by (6.3) that there is at least one orbit Xj , say, with the property

p o(Xj ). (6.4)

∈ X ∈ ={ }∈X \ −1 ={ } Let Y j where e Y (if Y y1,... j , then Y y1 e,... is in the same orbit as Y by deﬁnition), note that o(Y) = pr .NowJ is deﬁned by r J = stabG(Y ). By Lemma 5.6, J ≤ G, hence if we can show that o(J) = p , the result will follow. We do this by proving both pr ≤ o(J), and pr ≥ o(J). First, by the OrbitÐStabiliser Theorem (Theorem 5.7), we have r o(Xj ) = G : stabG(Y ) =[G : J ]=p m/o(J ), 6.2 Sylow Theory 121

using Lagrange’s Theorem for the ﬁnal equation. Now the last entity in this sequence of equations is an integer not divisible by p (by (6.4)), and so

pr | o(J), which implies pr ≤ o(J). (6.5)

For the reverse, note that J ⊆ YJ = Y (as e ∈ Y and J is the stabiliser of Y ). Hence o(J) ≤ o(Y) = pr , which with (6.5) proves the theorem.

3 Example We illustrate the above construction with the group D3 a,b | a = b2 = (ab)2 = e which has order 6, and we look for a subgroup of order 3. Here r = = 6 = p 3, m 2 and so the binomial coefﬁcient 3 20, and 3 20. Hence in this example X has 20 elements—the underlying set of D3 has 20 unordered 3-element subsets, and using the action deﬁned above these split into the four orbits given by:

{e,a,a2}, {b,ab,a2b}; {e,a,b}, {e,a2,ab}, {e,b,ab}, {a,a2,a2b}, {a,b,a2b}, {a2,ab,a2b}; {e,a,ab}, {e,a2,a2b}, {e,ab,a2b}, {a,a2,b}, {a,b,ab}, {a2,b,a2b}; {e,a,a2b}, {e,a2,b}, {e,b,a2b}, {a,a2,ab}, {a,ab,a2b}, {a2,b,ab}.

The ﬁrst orbit has an order not divisible by 3, and we can take its triple which contains e, that is, {e,a,a2} for Y and J . Now this stabiliser provides the subgroup of order 3 that we were looking for. Reader, ﬁnd a subgroup of order 2.

Second proof of Theorem 6.7 We use induction on o(G) = pr m, there is nothing to prove if o(G) = 1. We treat the cases of G non-Abelian, and G Abelian, separately. The ﬁrst of these has two subcases. Subcase 1.1 Z(G) < G and p | o(Z(G)). By Cauchy’s Theorem (Theorem 6.2) and as p divides o(Z(G)), Z(G) contains an element of order p, and so it contains a subgroup H of order p.By Problem 2.14(ii), H G, hence we can form the factor group G/H which has order pr−1m. By the inductive hypothesis, this factor group has a subgroup J/H of order pr−1, and the Correspondence Theorem now shows that J

r From this it follows that o(CG(yi)) = p n for some n satisfying 1 ≤ n

The third proof is as follows. By Problem 4.17, G is isomorphic to a subgroup of GLn(p) for some suitably chosen n. Also, by Problem 3.15(iii), we have ITn(p) is a n(n−1)/2 Sylow p-subgroup of GLn(p). Reader, check that o(ITn(p)) = p (there are p choices for each entry above the main diagonal) and use Theorem 3.15.Theresult follows by Problem 6.11(iv) if in this problem we put G = GLn(p), H = ITn(p) and J = G. For our first application of this result, we reprove Cauchy’s Theorem (Theo- rem 6.2), note that we did not use Cauchy’s result in the first (or third) proof of Theorem 6.7.Wehaveifo(G) = pr m where p m and r>0, then by the First Sylow Theorem, G has a subgroup P of order pr , and by Lagrange’s Theorem (Theorem 2.27), all elements of P have order a power of p.Ifh ∈ P and o(h) = ps , s−1 then hp has order p, as required by Cauchy’s result. We shall discuss some more applications and examples below, but first we make

Deﬁnition 6.8 Let o(G) = pr m where p is prime and p m. A subgroup of G with order pr is called a Sylow p-subgroup of G.ASylow subgroup is a Sylow p-subgroup for some unspeciﬁed prime p.

Note that a Sylow subgroup is only speciﬁed by its order. By Theorem 6.7,for every prime p, G has a Sylow p-subgroup P ;ifp o(G) then P =e.Alsoif m = 1, that is, G is a p-group, then G has only one Sylow p-subgroup—itself. We shall use the symbols P or Q for Sylow subgroups. If o(G) = pr m with p m and r>0 as above, then by Theorem 6.7, G always has at least one Sylow p-subgroup, and so by Theorem 6.5, G also has subgroup(s) of orders ps , for all s satisfying 1 ≤ s ≤ r. In the next theorem, we prove the converse, that is, every p-subgroup of G is contained in a Sylow p-subgroup of G. This theorem and its successor also provide more information and a number of further properties.

= = r1 ··· rk Examples (a) If G Cn where n p1 pk , then G has a unique normal (see ri = Theorem 6.10(iii) below) Sylow pi -subgroup of order pi , for each i 1,...,k,as all subgroups are normal in a cyclic group. This follows from Theorem 4.20. 2 (b) Consider the group A5 which has order 60 = 2 · 3 · 5. Theorem 6.7 implies that A5 has Sylow subgroups of orders 3, 4 and 5. In fact, A5 has ten subgroups of 6.2 Sylow Theory 123 order 3, ﬁve of order 4, and six of order 5 (Problems 3.3 and 6.13). But note that A5 has no subgroups of order 15, 20 or 30, see the example on page 101.

This second example shows that in the general finite case there is no extension of Sylow’s First Theorem to non-prime power divisors of the group order. There is an extension if the group is ‘soluble’, see Section 11.2. We come now to the remaining Sylow results. Let np denote the number of Sylow p-subgroups of G; by Theorem 6.7, np ≥ 1. The first of these theorems below gives the main properties including: For a fixed prime p, all Sylow p-subgroups of a group G are conjugate in G, and so in particular they are isomorphic.

Theorem 6.9 (Sylow Theorem, Part 2) Let G be a ﬁnite group. (i) Each p-subgroup of G is a subgroup of some Sylow p-subgroup of G. (ii) All Sylow p-subgroups of G are conjugate in G. (iii) np ≡ 1 (mod p).

Proof We use a similar method for all three parts of this theorem. Let P be a Sylow p-subgroup of G and let X denote the set of subgroups conjugate to P in G, that is, X ={g−1Pg : g ∈ G}. Further, let R be some p-subgroup of G and suppose R acts on X by conjugation: − − − − g 1Pg\x = x 1 g 1Pg x = (gx) 1P(gx) for g ∈ G and x ∈ R. (6.6)

As R is a p-group, the OrbitÐStabiliser Theorem (Theorem 5.7) shows that the order of an orbit of this action is a power of p, possibly p0 = 1. Now o(X ) = G : NG(P ) and p G : NG(P ) .

The equation here is given by Lemma 5.25(iii), and the non-divisibility property follows because P is a Sylow p-subgroup of both G and NG(P ) (they have the same (maximal) p-power order). Hence p o(X ) and so, as X is a disjoint union of its orbits (all of order a power of p), there must be at least one orbit of order 1(=p0). −1 Suppose {P1} is an orbit of the action (6.6) with order 1. Then x P1x = P1 for all x ∈ R,soP1R = RP1, and RP1 ≤ G (Problem 2.19(v)). But P1 ≤ RP1, hence o(P1) ≤ o(RP1). Applying Theorem 5.8 we have

o(RP1)o(R ∩ P1) = o(R)o(P1) or o(RP1) = o(P1)[R : R ∩ P1],

by Lagrange’s Theorem. As both P1 and R are p-groups (by deﬁnition and Problem 2.23), it follows that o(RP1) is a power of p, and so RP1 is a p-subgroup of G.AsP1 is a Sylow p-subgroup of G, and so its order is the maximum possible power of p. Hence

o(RP1) = o(P1) so RP1 = P1, which gives R ≤ P1.

This proves (i) because R is an arbitrary p-subgroup of G. 124 6 p-Groups and Sylow Theory

(ii) If R in the argument above is a Sylow p-subgroup of G, then o(R) = o(P1),soR = P1 where P1 is conjugate to P in G by the action defined above. This proves (ii), and it also shows that o(X ) = np. (iii) Suppose there is a second single-element orbit of the action (6.6), say {P2}. Then, taking R = P2, the first argument gives P1 = P2, that is there can only be one single-element orbit. As all other orbits have orders which are positive powers of p, (iii) follows because (by (ii)) o(X ) = np ≡ 1 (mod p). Notes Proposition (ii) in Theorem 6.9 is important for it implies that once we have found one Sylow p-subgroup P of a group, then all remaining Sylow p-subgroups will be isomorphic to P . This is not true for p-subgroups of smaller orders. For 3 example, the group S4 (note o(S4) = 2 · 3) has three Sylow 2-subgroups of order 8 each isomorphic to D4, but it has subgroups of order 4 isomorphic to both C4 and T2; see Chapter 8. Proposition (iii) implies that np = 0, and so Sylow p-subgroups exist, but we cannot use this to reprove the main theorem (Theorem 6.7) because we began the above proof by assuming the existence of a Sylow p-subgroup P . Proposition (i) holds for infinite groups, but note that there exist infinite groups (a) with non- isomorphic Sylow p-subgroups, and (b) with isomorphic Sylow p-subgroups which are not conjugate; see, for example, Suzuki (1982), page 191.

We collect together the remaining Sylow properties in the next theorem, they all follow easily from the results above.

Theorem 6.10 (Sylow Theorem, Part 3) Suppose o(G) = pr m where p m, and P is a Sylow p-subgroup of G.

(i) P is a normal Sylow p-subgroup of NG(P ). (ii) np =[G : NG(P )]. (iii) P G if and only if P is the unique Sylow p-subgroup of G, and P G implies P char G, see Problem 4.22. (iv) np | m.

Proof (i) By Lemma 5.25(i), P NG(P ).ButP is also a Sylow p-subgroup of NG(P ) as the orders of both groups have the same exponent of p in their prime factorisations, this gives (i). (ii) This follows directly from Theorem 6.9(ii) and Lemma 5.25(iii), or from the OrbitÐStabiliser Theorem (Theorem 5.7). (iii) If P is the unique Sylow p-subgroup of G, then by Theorem 6.9, all conjugates of P equal P , that is P G. Conversely, if P G then NG(P ) = G, and the result follows from (ii) and Theorem 6.9(ii). The second part is left as an exercise. (iv) We have m =[G : P ]= G : NG(P ) NG(P ) : P , by Lagrange’s Theorem and Problem 2.15, now use (ii). 6.2 Sylow Theory 125

We now restate the ﬁve main Sylow results which we shall often refer to as “Sylow 1”, ...,and“Sylow 5” (the numbering is not standard).

Five Main Sylow Theorems Suppose o(G) = pr m where p is prime, and p m. 1. G has at least one subgroup of order pr called a Sylow p-subgroup. 2. All Sylow p-subgroups of G are conjugate in G. 3. A Sylow p-subgroup P is unique if and only if P G. 4. If np denotes the number of Sylow p-subgroups of G, then

np ≥ 1,np | m and np ≡ 1 (mod p).

5. Each p-subgroup of G is contained in a Sylow p-subgroup of G.

Example We calculate the Sylow subgroups of S5; some more examples are given in Problem 6.12 and Chapter 8. 3 We note ﬁrst that o(S5) = 5!=2 · 3 · 5, and so S5 has Sylow subgroups of order 3, 5 and 8. Further, using Sylow 4 we have

n2 ≡ 1 (mod 2) and n2 | 15,

n3 ≡ 1 (mod 3) and n3 | 40,

n5 ≡ 1 (mod 5) and n5 | 24.

By Problem 3.3 and Theorem 2.29(ii), a non-neutral normal subgroup must have order at least 12, for by Problem 3.3 the conjugacy classes not containing the neutral element have orders between 10 and 24; and so n2,n3 and n5 are all larger than one by Sylow 3. Hence the third congruence above shows that n5 = 6, that is S5 has six (cyclic) subgroups of order 5. Note that S5 possesses 24 elements of order 5 and each subgroup isomorphic to C5 has four elements of order 5 and e; see Problem 2.5(iii). Secondly, the middle congruence above implies that n3 = 4, 10 or 40. By counting the number of elements of order 3 in S5 (there are 20), we see that n3 = 10 and the group has 10 (cyclic) subgroups of order 3. For the prime 2 we argue as follows. If we consider the subgroup generated by the permutations (1, 2, 3, 4) and (1, 3), we obtain an isomorphic copy of D4 in S5 (for a more detailed discussion of this point, see page 166). We have o(D4) = 8, hence this is a Sylow 2-subgroup and all Sylow 2-subgroups of S5 are isomorphic to D4. The first congruence above gives n2 = 3, 5 or 15. The correct value is 15 as we show now. The argument above can also be applied if we replace the two generators by either (1, 2, 4, 3) and (1, 4),or(1, 3, 2, 4) and (1, 2), this gives three copies of D4 in S5. Further, there are five ways of choosing four elements from five, and so the above can be repeated another four times giving a total of 15 Sylow 2-subgroups. The reader should list all these Sylow subgroups and also convince him(her)self that every subgroup of order 2 or 4 is contained in a Sylow 2-subgroup (Sylow 5). Note that S5 has 156 subgroups in 19 conjugacy classes, how many can you find? 126 6 p-Groups and Sylow Theory

6.3 Applications

Here we give a number of applications of the Sylow theory, see the comment at the end of this section and Web Section 6.5.

Groups whose Orders have at most Three Prime Factors

As a ﬁrst application of the Sylow theorems, we consider groups whose orders have a small number of (not necessarily distinct) prime factors. If o(G) is prime, then G is cyclic and has no non-neutral proper subgroups (Theorems 2.34 and 4.20). We show that if o(G) has two or three prime factors, then G has at least one proper non-neutral normal subgroup, and so is not simple. It can be shown that if o(G) has four prime factors, then G is simple only if o(G) = 60 and G A5 (Section 3.2), and if it has ﬁve prime factors then G is simple in a few well-known cases. (They are L2(7), L2(11) and L2(13) with orders 168, 660 and 1092, respectively, see Chap- ter 12 and the ATLAS 1985.) We begin by considering the two-prime case.

Theorem 6.11 If o(G) = pq, p and q are prime, and p ≤ q, then G has a normal (cyclic) subgroup of order q. If p = q it is unique.

It is also unique if p = q and G is cyclic, but it is not unique if G is elementary Abelian.

Proof If p = q this follows by Theorem 6.5, hence we may suppose p

nq ≡ 1 (mod q) and nq | p.

These conditions imply nq = 1, so by Sylow 3, P G and P is unique.

Corollary 6.12 If o(G) = 2p and p is an odd prime, then G is isomorphic to either C2p or Dp.

Proof By Theorem 6.11, G has a normal cyclic subgroup P of order p. Suppose P =a, and so ap = e. By Cauchy’s Theorem (Theorem 6.2), G also has an element b of order 2. As P G,wehave

b−1ab = bab = as,

for some positive integer s less than p.Ifr is also a positive integer, then bar b = (bab)(bab) ···(bab) = ars (as b2 = e) where there are r copies of bab in the second term. This proposition gives

2 a = b2ab2 = b(bab)b = basb = as , 6.3 Applications 127

and so s2 ≡ 1 (mod p) because a is an element of order p. This congruence has exactly two solutions: 1 and p − 1 (Appendix B). If s = 1, then G is Abelian, and so cyclic, see Chapter 7.Ifs = p − 1, then G is isomorphic to the pth dihedral group:

p 2 −1 p−1 G Dp = a,b | a = b = e,b ab = a .

The pq case, where p and q are both odd, is discussed in Problems 3.6, 6.14, 7.21, and 9.14, and in Web Section 14.3. For the three prime case we have

Theorem 6.13 If o(G) has exactly three prime factors, then G is not simple.

Proof There are three cases to consider. Case 1. o(G) = p3, p prime. In this case, the result follows by Theorem 6.5. Case 2. o(G) = p2q, p and q prime, and p = q.

We give two proofs. In the ﬁrst, we show that either np or nq equals 1 using Sylow 3 and 4. Hence suppose both np > 1 and nq > 1. As np | q and q is prime, we have np = q.Butnp ≡ 1 (mod p), and so q>p. (6.7)

2 We also have nq ≡ 1 (mod q) and nq | p ,so(6.7) and our supposition show 2 2 that nq = p . This implies that G has p cyclic subgroups each of order q (as q is prime), and so G has p2(q − 1) elements of order q, see Problem 2.5(iii). This further implies that G has exactly

p2q − p2(q − 1) = p2

elements whose orders are not equal to q. But this is impossible because we are assuming that G also has q subgroups each having order p2, and q ≥ 3by (6.7). Hence our supposition is false, either np = 1ornq = 1 (or both), and so G contains at least one normal subgroup by Sylow 3.

Second proof—this method has applications elsewhere. If p2 q and np > 1. This implies that G has distinct Sylow p-subgroups P and Q.LetR = P ∩ Q.By Theorem 5.8,wehaveo(R) = p (as o(P Q) ≤ p2q and P = Q). Now R is a p-group and a maximal subgroup of both P and Q. Using Problem 5.13(vi), this implies that

P,Q≤ NG(R), whichinturngives P,Q≤NG(R). But P is a maximal subgroup of G because it has prime index in G, this further implies that P,Q=NG(R) = G, and hence R G. This is impossible if G is simple, and so this case is established. 128 6 p-Groups and Sylow Theory

All three possibilities occur, for example, with groups of order 12, see Sec- tion 7.3. The cyclic group C12 has normal subgroups of order 3 and 4 (n2 = n3 = 1), D6 has a normal subgroup of order 3 (n2 = 3 and n3 = 1), and A4 has a normal subgroup of order 4 (n2 = 1 and n3 = 4); see Problem 6.13(i). Case 3. o(G) = pqr, p, q and r are prime, and p

As above suppose np,nq and nr are all larger than one. By Sylow 4, nr ≡ 1 (mod r) and nr | pq, hence nr = pq as r>q. Secondly, nq ≡ 1 (mod q) and nq | pr,sonq ≥ r as p

o(G) = pqr ≥ 1 + np(p − 1) + nq (q − 1) + nr (r − 1) ≥ 1 + q(p − 1) + r(q − 1) + pq(r − 1) = pqr + (q − 1)(r − 1),

which is impossible. Therefore, at least one of np,nq or nr is one, showing that G has at least one normal (Sylow) subgroup by Sylow 3.

Using this theorem the structure of all groups whose orders have at most three prime factors follows: For groups of order p2q, see Burnside’s classic 1897 (2nd edition, 1911 reprinted in 2004) textbook—the ﬁrst on group theory to be published in English. For groups of order pqr, see Theorem 6.18 (their Sylow subgroups are necessarily cyclic) and Problem 6.20. Groups of order 12 will be discussed at the end of the next chapter.

Frattini Argument and Nilpotent Groups

To prove our next result on so-called nilpotent groups, we need the Frattini Ar- gument, a simple procedure which has important applications in the theory; see Section 10.2. It also has one of the most elegant proofs in all mathematics!

Lemma 6.14 (Frattini Argument) If G is a ﬁnite group, K G, and P is a Sylow subgroup of K, then

G = NG(P )K.

Proof For g ∈ G we have

g−1Pg ⊆ g−1Kg = K,

as P ≤ K G. Hence both P and g−1Pg are Sylow subgroups of K (they have the same order, Problem 2.23), and so by Sylow 2 they are conjugate 6.3 Applications 129

in K. Therefore, we can ﬁnd k ∈ K to satisfy k−1 g−1Pg k = (gk)−1P(gk) = P.

By Deﬁnition 5.24, this shows that gk ∈ NG(P ), and so

−1 g ∈ NG(P )k ⊆ NG(P )K ⊆ G.

The result follows because this argument applies to all g ∈ G.

Example Suppose G = S5, K = A5 and P =(1, 2, 3); see the example on page 125.NowP is a Sylow 3-subgroup of K and K G, and so Frattini’s Ar- gument gives S5 = NG(P )A5 which shows that NG(P ) must contain an odd permutation. In fact, NG(P ) =(1, 2, 3), (2, 3), (4, 5) D6, a group of order 12 (look for the permutations in S5 which commute with (1, 2, 3)); see Section 7.3.

As an illustration of this result we prove the following results, two of many dealing with properties of the normaliser; see Problem 6.10.

Corollary 6.15 If P is a Sylow p-subgroup of a ﬁnite group G, and NG(P ) ≤ H ≤ G, then H is self-normalising, that is, H = NG(H ).

Note this shows that NG(P ) is itself self-normalising.

Proof We have by hypothesis and Theorem 6.10(i)

P NG(P ) ≤ H NG(H ) ≤ G,

and P is a Sylow p-subgroup of each group in this sequence. Applying the Frattini Argument with NG(H ) for G, and H for K,wehave

= NG(H ) NNG(H )(P )H. ≤ ≤ ⊆ = But NNG(H )(P ) NG(P ) H ,asNG(H ) G, and so NG(H ) H .

A finite group is called nilpotent if all of its Sylow subgroups are normal. A second definition is: Every maximal subgroup of G is normal in G (cf. Theorem 6.6). Here we use the Frattini Argument to show that this second property implies the first. In the next chapter (page 144), we prove the reverse implication and introduce a third equivalent definition involving direct products; a full discussion of nilpotency is given in Chapter 10.

Theorem 6.16 If every maximal subgroup of a ﬁnite group G is normal in G, then every Sylow subgroup of G is normal in G; that is, G is nilpotent. 130 6 p-Groups and Sylow Theory

Proof Suppose P is a non-normal Sylow subgroup of G. By Lemma 5.25,

P ≤ NG(P ) < G.

Hence there is a maximal subgroup H of G (possibly NG(P ) itself) which satisﬁes

P ≤ NG(P ) ≤ H

G = NG(P )H ≤ H,

which is impossible because maximal subgroups are always proper, and H is maximal by deﬁnition. Therefore, P G. This applies to all Sylow subgroups of G and so the result follows.

Frattini introduced his argument to show that the subgroup of a group formed by the intersection of all of its maximal subgroups, the so-called Frattini subgroup, is nilpotent; we prove this result in Chapter 10 where further applications are given. Web Section 6.5 gives some more applications of Sylow’s theorems. Two are quite important and are used in a few of the problems, hence we shall state them here; their proofs (the second is basically a consequence of the ﬁrst) are given in the web section. The ﬁrst—one of Burnside’s most important—concerns normal complements and is as follows.

Theorem 6.17 (Burnside’s Normal Complement Theorem) Suppose G is a ﬁnite group with Sylow subgroup P . If CG(P ) = NG(P ), then P has a normal complement K in G, that is, G has a normal subgroup K with the property G = PK.

In Problem 5.17(ii), we showed that the centraliser/normaliser condition in this theorem is equivalent to the condition P ≤ Z(NG(P )). The second application, which is due to Hölder, Burnside and Zassenhaus, characterises groups with cyclic Sylow subgroups.

Theorem 6.18 Suppose G is a ﬁnite group all of whose Sylow subgroups are cyclic, then G has a presentation in the form

a,b | am = bn = e,b−1ab = ar where m is odd, (m, n(r − 1)) = 1, 0 ≤ r

Groups satisfying these conditions are called Frobenius or metacyclic,theyare denoted by Fm,n,orFm,n,r if for ﬁxed m and n there are several values of r (and so 6.4 Problems 131 several non-isomorphic groups). For example, there are six non-isomorphic groups of order 42, all Frobenius; see Web Section 6.5 and Problem 8.3.

6.4 Problems

Problem 6.1 (i) Complete the proof of Theorem 6.6 using Cauchy’s (Theo- rem 6.2) and the Correspondence (Theorem 4.16) Theorems. (ii) Using Theorem 4.20, show that every non-generator in a cyclic p-group is a p-th power. (Hint. See Problem 2.13(ii).) (iii) Use the Class Equations (Theorem 5.20) to show that if G is a ﬁnite p-group, K G and o(K) = p, then K ≤ Z(G). This result can also be derived using Prob- lem 5.3.

p Problem 6.2 Given a group G,letGp ={g ∈ G : g = e}.

(i) Prove that Gp char G if G is an Abelian p-group; see Problem 4.22. (ii) Let G be a p-group and, for this problem, let Exp(G) denote the subgroup generated by the elements of the set Gp. Note that the inverse of an element of order p is itself an element of order p. Use Problem 4.6 and one of the commutator properties given in Problem 2.17 to show that (a) Exp(G) = Gp if p>2, and (b) G/ Exp(G) is Abelian.

Problem 6.3 For each group G of order 8, see Section 6.1, list their subgroups, ﬁnd Z(G), and determine the isomorphism type of G/Z(G) using direct calculation and some results from this chapter and its predecessor. Also list their normal subgroups and draw subgroup lattice diagrams.

2 −3 Problem 6.4 (An Example on 2-Groups) Let G1 =a,b | b = e,bab = a and 2 3 G2 =a,b | b = e,bab = a .

8 (i) Show that a = e in each case, and so prove that both G1 and G2 have order 16. (ii) Find the subgroups of G1 and G2, and determine which are normal. (iii) Verify all parts of Theorems 6.5 and 6.6 for both G1 and G2. (iv) Find the centres and the derived subgroups. (v) Show that the subgroup lattice diagrams for G1 and C8 ×C2, see Chapter 7,are identical except for the ‘top’ group. Hence prove that non-isomorphic groups of the same order can have identical proper subgroup lattices.

The group G2 is called semi-dihedral, another representation is given in Prob- lem 6.23. The reader should note the similarities and differences between this group and the dihedral group with the same order, that is D8, and see Problem 8.12.Also note that the group G1 can be represented by the semi-direct product C2 C8 which is discussed in Chapter 7. 132 6 p-Groups and Sylow Theory

Problem 6.5 (Extra-Special Groups) For more details on these groups, see Aschbacher (1994), Chapter 2. (i) Let G be a non-Abelian group of order p3 where p is an odd prime, see Prob- lem 5.22(iii). By considering G/Z(G), show that there are at most two isomorphism classes for G, and so deduce that G has one of the following presentations

= | p2 = p = = 1+p ES1(p) a,b a b e,ab ba , p p p ES2(p) = a,b,c | a = b = c = e,c =[a,b],ca= ac,cb = bc . Each of these groups is called extra-special, that is, they satisfy the conditions: Z(G) is cyclic and Z(G) = G = (G); see Section 10.2. Note that for the second group every non-neutral element has order p, but it is not elementary Abelian if p>2. (If p = 2 the group is elementary Abelian, see Corollary 2.20.) (ii) Show that ES2(p) is represented by the matrix group IT3(p); see Problem 3.15. We shall give a representation of ES1(p) in Problem 7.22. (iii) Find the subgroups of these groups when p = 3, and indicate which are normal and which are maximal.

Problem 6.6 Suppose G is a ﬁnite p-group with a single subgroup of order p, and if p = 2 suppose also G is Abelian. Show that G is cyclic using the following method. Use induction on n where o(G) = pn.Forn = 2 use Theorem 5.22, and for n = 3 use the previous problem and Section 7.2. In the general case, let K ≤ G with o(K) = p, and suppose G/K has two subgroups H/K and J/K both with order p. For more details, see Doerk and Hawkes (1992), page 204. Show ﬁnally that the extra condition in the case p = 2 is necessary.

Problem 6.7 (The group Cp∞ —An Inﬁnite Group all of whose Proper Subgroups are Finite) Fix a prime p. The group Cp∞ has the following presentation. The generators are a0,a1,..., and the relations are given by

aiaj = aj ai, for all i and j, p = p = p = a0 e, a1 a0, ..., an+1 an,...

(i) Prove that Cp∞ is an inﬁnite Abelian p-group. (ii) Show that if H ≤ Cp∞ and am ∈ H , then an ∈ H for all n

Problem 6.8 (i) For the given groups and primes determine the isomorphism classes of the Sylow p-subgroups of (a) S6 with p = 3, (b) A6 with p = 2, and (c) SL2(5) again with p = 2. (Hint. Use Problem 3.19(i) for (c).) 6.4 Problems 133

(ii) Prove that if n is odd, then all Sylow subgroups of Dn are cyclic. Is this true if n is even? (iii) Show that Sylow subgroups can intersect both neutrally and non-neutrally. To be more precise, there are groups G with distinct Sylow p-subgroups P1,P2 and P3 (with p ﬁxed) that satisfy:

P1 ∩ P2 =e and P1 ∩ P3 > e.

An example can be found in a subgroup H of S6 of permutations in which the elements of the subsets {1, 2, 3} and {4, 5, 6} are permuted amongst themselves but there is no mixing; that is, H S3 × S3, see Problem 3.9 and Section 7.1.

Problem 6.9 (Properties of Sylow Subgroups 1) (i) Suppose H ≤ G, P is a Sylow p-subgroup of H , and Q is a Sylow p-subgroup of G with the property P ≤ Q, prove that P = Q ∩ H . This result can be extended to: If H ≤ G and G has a Sylow p-subgroup P , then for some g ∈ G we have H ∩ g−1Pg is a Sylow p-subgroup of H , and this extension can be used as the starting point for another proof of the First Sylow Theorem (Theorem 6.7). See also Problem 6.10(iv). (ii) Show that if K is a normal p-subgroup of a ﬁnite group G, then K ≤ P for all Sylow p-subgroups P of G. (iii) Let H be a non-Sylow p-subgroup of G,soo(H) is a power of p and p |[G : H ]. Using a coset action and Problem 5.3 show that H

Problem 6.10 (Properties of Sylow Subgroups 2) Suppose G is ﬁnite and P is a Sylow p-subgroup.

(i) Show that if NG(P ) ≤ H ≤ G, then [G : H]≡1 (mod p). (ii) Show that if g and h belong to CG(P ) and are conjugate in G, then they are also conjugate in NG(P )—a useful fact. (iii) Using Theorem 5.8, prove that if K G, then K ∩P is a Sylow subgroup of K, and PK/K is a Sylow subgroup of G/K. (iv) Give an example to show that if J is a non-normal subgroup G, then J ∩ P need not be a Sylow subgroup of J ; see (iii). (v) Use (iii) to show that if K G, then np(G/K) ≤ np(G). 134 6 p-Groups and Sylow Theory

Problem 6.11 (Properties of Sylow Subgroups 3) Let H ≤ G.

(i) Let {p1,...,pk} be a list of the prime divisors of o(G) and let Pi be a Sylow pi -subgroup of G for i = 1,...,k. Show that G =P1,...,Pk and G = P1P2, if k = 2. See also Condition 6 on page 242. (ii) If K1,K2 G and P is a Sylow subgroup of G, show that

K1K2 ∩ P = (K1 ∩ P )(K2 ∩ P) and K1P ∩ K2P = (K1 ∩ K2)P.

(iii) Let P and Q be distinct Sylow p-subgroups of H , and let P1 ≥ P and Q1 ≥ Q where P1 and Q1 are Sylow p-subgroups of G. Prove that P1 and Q1 are also distinct. Use this to show that np(H ) ≤ np(G). (iv) Let J ≤ G and let H be a Sylow p-subgroup of G. Show how to ﬁnd an element a ∈ G with the property: J ∩ a−1Ha is a Sylow p-subgroup of J . (Hint. Use Problem 2.29.)

Problem 6.12 (i) If P is a Sylow p-subgroup of the symmetric group Sp, show that = − o(NSp (P )) p(p 1); see Problems 3.11 and 6.19. (ii) Let G be a transitive subgroup of Sp (that is for each pair of integers r and s in {1,...,p} there exists a permutation in G which maps r to s;see page 94), let np equal the number of Sylow p-subgroups in G as usual, and let P , a Sylow p-subgroup of G, be one of these subgroups. Show that if np > 1, then P is not self-normalising (that is, P

Problem 6.13 (i) Show that if o(G) = 12, then G has a normal Sylow subgroup of order 3 or 4. (ii) Suppose m is a proper divisor of 60 which is itself divisible by 5. Prove that if o(G) = m, then G has a unique Sylow 5-subgroup. (iii) List the Sylow subgroups of A5.

Problem 6.14 Let G be a non-Abelian group of order pq where p and q are primes, and p | q − 1. Using Theorem 6.11, (a) ﬁnd a presentation for G with two generators, (b) give a formula in terms of the generators of the product of one element by another, and (c) show that up to isomorphism there is only one group satisfying these conditions. See also Problems 3.6, 7.21 and 9.14, and Web Section 15.3.

Problem 6.15 By applying the results proved in this chapter and its predecessor including Theorem 5.15, show that the smallest non-Abelian simple group has order 60. Note that the next non-Abelian simple group has order 168, try to prove this; some parts are covered in Problem 6.17. 6.4 Problems 135

Problem 6.16 (i) Using the method indicated below, or otherwise, show that if G has order 60 and possesses more than one Sylow 5-subgroup, then it is simple. Hence show that A5 is simple. Method: suppose K G and 5 | o(K). Use Prob- lems 6.13(ii) and 4.22 to show this is impossible. Deduce 5 | o(G/K), and so prove that o(K) = 12. Obtain the ﬁnal contradiction using Problem 6.13(i). (Hint. You need to show that if H is a Sylow 5-subgroup of G, then HK = G.) (ii) Prove that a simple group G of order 60 is isomorphic to A5. One method is as follows. (a) Show that G has a subgroup of index 5 by considering Sylow 2-subgroups and their intersections, supposing the contrary, and counting elements of order a power of 2. (b) Apply Theorem 5.15 to show that G is isomorphic to a subgroup of S5, and then use Theorem 5.8. Deduce A5 L2(4) L2(5) using Theorems 3.11 and 12.7. Note that L2(q) = SL2(q)/Z(SL2(q)), see Section 12.2.

Problem 6.17 Using one of the methods suggested, or otherwise, show that there are no simple groups with the following orders. (i) 90. Method—by the Sylow theory if the group G is simple, then n3 = 10 and n5 = 6. Firstly, use a counting argument to show that each pair of the ten Sylow 3-subgroups cannot have neutral intersection. Secondly, if P and Q are distinct Sy- low 3-subgroups of G with non-neutral intersection, R = P ∩ Q and S = NG(R), show that o(S) = 18, 45 or 90 applying the method used in the second proof of Theorem 6.13, Part 2. Finally, show that each of these cases is impossible if G is simple. Using Problem 3.7(ii), this fact can also be established by defining an injection from the supposed simple group into A6, and arguing that A6 has no subgroups of index 4. (ii) 108. Method—use Theorem 5.15. Similar methods apply for groups of order pr (p + 1) if p is prime and r>1. (iii) 112. The method suggested works for groups G of order pnq. Assum- ing that G simple, show first that np = q. Prove that if all pairs of distinct Sylow p-subgroups P and Q have neutral intersection, then using a counting argument, deduce nq = 1 and so obtain a contradiction. Now suppose R = P ∩ Q where o(R) is of maximal size which we assume to be larger than 1, and let S = NG(R).By Problem 5.13(vi), note that R is a proper subgroup of both NP (R) and NQ(S). Secondly, suppose S has a single Sylow p-subgroup T , say, and let U be the Sy- low p-subgroup of G containing T . Show that P = U = Q, and so deduce T is not unique. Use this to show that S has q Sylow p-subgroups, and by applying Problem 6.10(iii) deduce that all of these subgroups contain R. Now use Prob- lem 6.10(iii) again, and the supposed simplicity of G, to show that R =e.As in the case of groups of order 90, this non-simplicity result (for a group of order 112) can also be established by considering injections into the alternating group A7. (iv) 132. Method—use the Sylow theory and a counting argument. (v) 144. Method—assume that n3 > 1, so n3 = 4 or 16. Apply Theorem 5.15 in the first case, and in the second case, use a counting argument to show that if each pair of Sylow 3-subgroups has neutral intersection, then n2 = 1. If not, apply the first method suggested in (i) treating the normalisers of orders 18, 36, 72, and 144 separately. 136 6 p-Groups and Sylow Theory

Problem 6.18 Prove that Aut Q2 S4; one method is as follows. Noting that automorphisms map maximal subgroups to maximal subgroups (this is part of the Correspondence Theorem (Theorem 4.16)), use Theorem 5.15 to construct a homomorphism φ : Aut Q2 → S3, and show that φ is surjective (these automorphisms interchange elements of order 4). Secondly, use Corollary 5.27 to show that there are four inner automorphisms (each of these permute the elements of the maximal subgroups but leave the subgroups themselves unaltered); thirdly, deduce o(Aut Q2) = 24, and so prove the result using Section 8.1.

Problem 6.19 Working in S5,letσ = (1, 2, 3, 4, 5), P =σ ≤S5 and K = = = NS5 (P ). Show that o(K) 20 and K σ,τ where τ (2, 3, 5, 4); see Prob- lems 3.11 and 6.12.

Problem 6.20 (i) Using the Frattini Argument (Lemma 6.14), show that if H G, P is a Sylow subgroup of G and P H , then P G; note that normality is not transitive in general. (ii) Suppose G has square-free order. You are given: G is not simple, this can be proved using Burnside’s Normal Complement Theorem (Theorem 6.17). Show that if p∗ is the largest prime factor of o(G), then G has a normal p∗-subgroup by applying the following method: use induction on the number of factors of o(G), Theorem 6.11, and Problem 6.10(iii). If H is the given normal subgroup of G, treat the cases (a) p∗ | o(H), and (b) p∗ o(H) separately, then use (i). See also Theo- rem 6.18.

Problem 6.21 (i) Let K1,K2 G have the property: K1K2 = G, and let H ≤ G. Show that H = (K1 ∩ H )(K2 ∩ H). (Hint. Use Problem 2.18.) (ii) Secondly, show by an example that the result in (i) is false if either K1 or K2 is not normal; but see (iv) below. In one example, G = S4. (iii) Suppose K G, P is a p-subgroup of G, and (o(K), p) = 1. Show that

NG/K (P /K) = NG(P )/K.

(Hint. Use the Frattini Argument (Lemma 6.14).) (iv) Now assume that H is a complement of K in G,soG = HK where H ≤ G. Using (i) and the Second Isomorphism Theorem (Theorem 4.15) show that NG(P ) = K ∩ NG(P ) H ∩ NG(P ) .

Problem 6.22 Suppose P1,...,Pm is a list of the Sylow p-subgroups of a ﬁnite group G. Note that m ≡ 1 (mod p). Further, let S(P ) denote the group of permutations of the set {P1,...,Pm}, and deﬁne a map θ : G → S(P ) so that gθ is the −1 permutation that maps Pi to g Pig,fori = 1,...,m. (i) Show that θ is a homomorphism and determine its kernel. (ii) Find the kernel of θ when G = Dn, n is odd, and p = 2. Hence show that Dn is isomorphic to a subgroup of Sn when n is odd. 6.4 Problems 137

Problem 6.23 (Project—Subgroups of GL2(3)) In this project, you are asked to ﬁnd all 55 subgroups of the matrix group GL2(3) some of which are semi-dihedral,see Problem 6.4. This project is a continuation of the one given in Problem 3.24 where you were asked to show that the group GL2(3) has the following presentation

a,b,c | a8 = b2 = c3 = e,bab = a3,bcb= c2,c2a2c = ab,c2abc = aba2 .

You were further asked to find matrix and permutation representations. As noted there the work can be done by hand, but a computer package which includes matrix calculations modulo a prime (3 in this case) would be an asset. Method: To find the subgroups, use some results from Section 3.3 to write out all elements of the group with their orders. Secondly, using Problem 5.19 show that the elements of the centre together with those of order 4 form a normal subgroup K isomorphic to the quaternion group Q2. Thirdly, show that the subset of elements of the group whose orders are a power of 2 define the Sylow 2-subgroup(s) and, using Problem 6.9(ii), show that K is a subgroup of each of them. Fourthly, determine these Sylow 2-subgroups. Now use Problem 5.19 again to find the subgroups of order 12, note there are four of them, and using suitably chosen elements of the Sylow 2-subgroup(s) find the remaining subgroups (seven in all) with order 8. Finally, using the methods applied in the last part of Section 6.1, find the cyclic and remaining dihedral subgroups of the group, indicate which subgroups are normal, and find the centre and the derived subgroup. Chapter 7 Products and Abelian Groups

This chapter can be read before Chapters 5 and 6, see the note on page 91. Suppose (Gi, i), i = 1, 2,..., are groups, and H is the (set-theoretic) Cartesian product of their underlying sets (page 278) H = G1 × G2 ×···= Gi. i

A number of new groups can be formed using H as the underlying set. The operations of these new groups are constructed using the operations i of Gi , i = 1, 2,.... For example, if G1 = C2 and G2 = C3, then H is the set of pairs (a, b) where a ∈ C2 and b ∈ C3, and we deﬁne a new operation by

(a1,b1) (a2,b2) = (a1 1 a2,b1 2 b2).

It is easily seen that this construction defines a group which we denote by C2 ×C3.It is called the direct product of C2 and C3; direct because there is no mixing between the first and second arguments. In the first argument, we have elements ai ∈ G1 and we use the operation 1, and in the second we have bi ∈ G2 and use the operation 2. In other products, some mixing is allowed, the terms in one or both arguments of the product are constructed using ai , bi, 1 and 2, see Section 7.3. In this chapter, we discuss direct products, and two types of product where some mixing is allowed, they are called semi-direct and wreath; the latter, a special case of the former, is only considered briefly. We have seen some examples in the previous chapters; there is no circularity here because the prerequisites for most of the work in this chapter were given in Chapter 2, and Sections 4.1 and 4.2. We shall also show that all finite Abelian groups can be represented as direct products of cyclic groups—an important result with applications to other areas of mathematics. Some examples of groups built up using the semi-direct construction will be given in the final section. A brief account of the basic facts about infinite Abelian groups can be found in Web Section 7.5. H.E. Rose, A Course on Finite Groups, 139 Universitext, DOI 10.1007/978-1-84882-889-6_7, © Springer-Verlag London Limited 2009 140 7 Products and Abelian Groups

7.1 Direct Products

Given sets Gi , i = 1,...,n, the underlying sets of the groups Gi , we can form the (set-theoretic) Cartesian product of ordered n-tuples of elements of G1 to Gn by G1 ×···×Gn = (a1,...,an) : ai ∈ Gi,i = 1,...,n , see Appendix A. This will be used as the starting point for the construction of new groups. The (ﬁnite) direct product is given by

Deﬁnition 7.1 For each i = 1,...,n,let(Gi, i) be a group with neutral element ei .OnthesetG = G1 ×···×Gn as given above, deﬁne the operation by

(a1,...,an) (b1,...,bn) = (a1 1 b1,...,an n bn), where ai,bi ∈ Gi ,fori = 1,...,n. We denote the set G with the operation by G1 ×···×Gn, it is a group called the direct product of G1,...,Gn; see Theo- rem 7.2. Each Gi is called a factor of the product group G.

Note that we use the same symbol × for the set-theoretic Cartesian product and the group-theoretic direct product.

Theorem 7.2 If Gi are groups for i = 1,...,n, then the set G1 ×···×Gn, with the operation given in Deﬁnition 7.1, also forms a group.

Proof The operation is clearly closed and associative because the constituent operations i have these properties. The neutral element is

(e1,...,en),

and the inverse operation is given by −1 = −1 −1 (a1,...,an) a1 ,...,an , using the inverse operations in the constituent groups.

The group constructed in Theorem 7.2 is sometimes called the external direct product of the groups Gi . We are not restricted to finite products, products with an infinite number of factors can be defined similarly; see Web Section 7.5. We shall give some examples later, but first we derive the basic properties. To ease the notation, we give the results for the product of two groups first, they can easily be extended to products with a finite number of factors; see Lemma 7.5 and Theo- rem 7.6.

Lemma 7.3 Suppose H and J are groups. (i) If o(H),o(J) < ∞ then, o(H × J)= o(H) · o(J). 7.1 Direct Products 141

(ii) H and J are Abelian if and only if H × J is Abelian. (iii) H ×e H × J , e×J H × J , H ×eH , and e×J J . (iv) (H × J)/(H ×e) J and (H × J)/(e×J) H .

Proof (i) and (ii) These are immediate consequences of the deﬁnitions. (iii) and (iv) Deﬁne a map φ : H × J → J by

(h, j)φ = j for h ∈ H, j ∈ J.

It is easily checked that this deﬁnes a surjective homomorphism with kernel H ×e. The ﬁrst parts of both (iii) and (iv) follow by the First Isomorphism Theorem (Theorem 4.11), and the second parts follow in the same way. Simi- lar arguments can be applied to establish the third and fourth parts of (iii), the reader should write them out.

Secondly, we consider these products from the opposite point of view: Under what conditions is a group isomorphic to a direct product of two or more of its subgroups? For example, we constructed the group C2 × C3 on page 139, but in this case we obtained nothing new, for this group is isomorphic to C6; that is, C6 can be treated as a direct product of its two subgroups isomorphic to C2 and C3,see Theorem 4.20. The next result gives necessary and sufﬁcient conditions in the two- subgroup case.

Theorem 7.4 Suppose H and J are subgroups of the group G, H G, J G, H ∩ J =e, and HJ = G. (i) If g = hj where g ∈ G, h ∈ H and j ∈ J , then h and j are uniquely determined by g. (ii) hj = jh, for all h ∈ H and j ∈ J . (iii) G H × J J × H .

In this case, we say that G is the internal direct product of its subgroups H and J .

Proof Throughout we assume that h, h ,h1,h2 ∈ H and j,j ,j1,j2 ∈ J . (i) As G = HJ, for each g ∈ G we can ﬁnd h ∈ H and j ∈ J to satisfy

g = hj .

Proposition (i) follows for if g = hj = h j , then h −1h = j j −1 where h −1h ∈ H and j j −1 ∈ J .ButH ∩ J =e, and so h −1h = j j −1 = e which gives h = h and j = j . (ii) As H G,wehavej −1h−1j ∈ H which shows that j −1h−1jh∈ H ; also as J G,wehavej −1h−1jh ∈ J . But as above H ∩ J =e, hence j −1h−1jh= e,orhj = jh, which gives (ii). 142 7 Products and Abelian Groups

(iii) We deﬁne a map ψ : G → H × J by

if g = hj then gψ = (h, j).

The uniqueness condition proved in (i) shows that ψ is well-deﬁned, and it is clearly surjective (as (hj)ψ = (h, j)). To prove injectivity we argue as follows. Suppose g1ψ = g2ψ, where gi = (hi,ji),fori = 1, 2. This gives (h1,j1) = (h2,j2), and so = −1 = −1 −1 (e, e) (h1,j1)(h2,j2) h1h2 ,j1j2 ,

which is only possible if h1 = h2 and j1 = j2. But then g1 = g2 proving injectivity. For the homomorphism property, we have

g1ψg2ψ = (h1j1ψ)(h2j2ψ)= (h1,j1)(h2,j2)

= (h1h2,j1j2) = h1h2j1j2ψ

= h1j1h2j2ψ = g1g2ψ,

using the deﬁnition of ψ and (ii) for the last but one equation (that is, h2j1 = j1h2). Putting these properties together, we see that ψ is an isomorphism. The second isomorphism in (iii) is an direct consequence of (ii).

The converse of Theorem 7.4 is given by Lemma 7.3. The n-subgroup versions of Lemma 7.3 and Theorem 7.4 follow, their proofs will be left as exercises for the reader, see Problem 7.1. Note that (iii) and (iv) show that the direct product is both associative and commutative.

Lemma 7.5 Suppose H1,...,Hn are groups.

(i) If o(Hi)<∞ for i = 1,...,n, then o(H1 ×···×Hn) = o(H1) ···o(Hn). (ii) H1,...,Hn are all Abelian if and only if H1 ×···×Hn is Abelian. (iii) If J,K and L are direct products of groups (with one or more factors), then J × (K × L) (J × K)× L. { } { } ×···× ×···× (iv) If i1,...,in is a permutation of 1,...,n , Hi1 Hin H1 Hn. ∗ ×···× × × ×···× (v) For each i, let Hi denote the group e e Hi e e with n factors and where Hi is in the ith place, then

∗ ×···× Hi H1 Hn, and ×···× ∗ ×···× × ×···× (H1 Hn)/Hi H1 Hi−1 Hi+1 Hn.

Theorem 7.6 If H1,...,Hn are subgroups of a group G, then G H1 ×···×Hn if and only if the following three conditions hold:

(i) Hi G, for i = 1,...,n, (ii) G = H1 ···Hn, (iii) Hi ∩ H1 ···Hi−1Hi+1 ···Hn =e, for i = 1,...,n. 7.1 Direct Products 143

Condition (iii) in this theorem cannot be replaced by

Hi ∩ Hj =e, for 1 ≤ i

t u Example Let G =a b : t = 0, 1, u = 0, 1,the4-group T2. It has three subgroups of order 2: H1 =a,H2 =b and H3 =ab; and Hi G for i = 1, 2, 3, G = H1H2H3, and H1 ∩ H2 = H1 ∩ H3 = H2 ∩ H3 =e.ButG H1 × H2 × H3 because G has order 4 whilst H1 × H2 × H3 has order 8.

Uniqueness of Representation

Suppose G is represented as a direct product of (some of) its subgroups H1,...,Hm. We can ask under what conditions is this representation unique? Clearly, it is not unique if some of the subgroups Hi are themselves direct products. So we ask: If

G H1 ×···×Hm, and each Hi is indecomposable, that is, it cannot be represented as a direct product with non-neutral factors then, apart from the order of the terms, is this representation unique? The answer is yes in the finite and some infinite cases. This result was first conjectured by J.H.M. Wedderburn (1882Ð1948), and it was proved by him, Remak, Krull and Schmidt. For the proof, and details of the infinite case, the reader should consult Suzuki (1982), page 127ff. We shall consider the Abelian case in the next section. For our first application of the results above, we return to the topic of cyclic groups; see Section 4.3.Wehave

= s1 sk Theorem 7.7 (i) If n p1 ...pk is the prime factorisation of n, then

Cn C s1 ×···×C sk . p1 pk

(ii) If (m, n) = 1 then Cmn Cm × Cn.

= si = Proof (i) Let ni n/pi ,fori 1,...,k, and let g be a generator of the n n cyclic group Cn. Then g i generates a subgroup Pi =g i of Cn of order si pi . (Incidentally, Pi is the unique Sylow pi -subgroup of Cn; one proof of Sylow’s First Theorem reduces the general case to this example; see Alperin and Bell 1995, page 64.) By unique factorisation of n,wehaveCn = P1 ···Pk. Also Pi ∩ P1 ···Pi−1Pi+1 ···Pk =e,fori = 1,...,k as Pi only contains elements of order a power of pi . Now apply Theorem 7.6 as all subgroups are normal. (ii) This is an immediate consequence of (i) and Lemma 7.5. 144 7 Products and Abelian Groups

For our second application, we consider subgroups of direct product groups. We have the following useful

Lemma 7.8 If G H1 ×···×Hk, (o(Hi), o(Hj )) = 1 when i = j, and J ≤ G, then

J (H1 ∩ J)×···×(Hk ∩ J).

Proof We treat the case k = 2, the general case follows by induction. The second hypothesis gives (H1 ∩J)∩(H2 ∩J)=e. Also, using direct product properties, and Problems 2.5 and 2.14,wehaveHi ∩ J J for i = 1, 2. Hence we can construct the direct product (H1 ∩ J)× (H2 ∩ J)inside J . By the ﬁrst hypothesis, if j ∈ J , then j = h1h2 where hi ∈ Hi for i = 1, 2. The result will follow if we show that h1,h2 ∈ J as j is an arbitrary element of J . As (o(h1), o(h2)) = 1, and h1 and h2 commute (Theorem 7.4(i)), we have o(h1h2) = o(h1)o(h2), and by Theorem 7.7(ii), h1h2h1×h2. Hence j=h1h2h1×h2, and so h1,h2 ∈j≤J . The lemma follows.

Examples (a) Let G = C3 × D4, see Appendix C. The group G has a subgroup J isomorphic to C12, and so the lemma gives J J ∩ C3 × J ∩ D4, that is, J ∩ C3 C3, J ∩ D4 C4, and J C12 C3 × C4. (b) Referring back to the example given below Theorem 7.6, suppose G, H1 and H2 are as deﬁned there, and we set J = H3.NowJ ≤ G and H1 ∩J = H2 ∩J =e, and so the conclusion of Lemma 7.8 is false in this case. But o(H1) = o(H2) = 2, and so this shows that second condition in the lemma above is essential.

We introduced finite nilpotent groups in Section 6.3 and we shall discuss general groups of this type in Chapter 10. In the finite case, there are a number of equivalent definitions one of which involves direct products (Theorem 10.9), and so we prove the following result now.

Theorem 7.9 Suppose G is a ﬁnite group. The statements below are equivalent: (i) G is the direct product of its non-neutral Sylow subgroups. (ii) All maximal subgroups of G are normal in G. (iii) All Sylow subgroups of G are normal in G.

Note that if G is a p-group, the result follows from Theorem 6.6. Also, using the Frattini Argument (Lemma 6.14), we have shown previously that (ii) implies (iii), see Theorem 6.16. It is also of interest to note the range and number of subsidiary results that are used in the proof given below.

= s1 ··· sk = Proof Let o(G) p1 pk and, for i 1,...,k,letPi be a Sylow pi -subgroup of G. We need to show that (i) implies (ii), and (iii) implies (i). 7.1 Direct Products 145

(i) implies (ii) By (i) we have

G P1 ×···×Pk.

By Theorem 7.6, this shows that Pi G,fori = 1,...,k, and so Pi is the unique Sylow pi -subgroup of G by Theorem 6.10.LetH be a maximal subgroup of G.As(o(Pi), o(Pj )) = 1ifi = j, we can apply Lemma 7.8 to obtain, for some j satisfying 1 ≤ j ≤ k,

H (P1 ∩ H)×···×(Pj ∩ H)×···×(Pk ∩ H)

P1 ×···×(Pj ∩ H)×···×Pk,

where Pj ∩H

H<(P1 ∩ H)× P2 ×···×Pk

but this is impossible as it contradicts the maximality of H in G. For the same reason, Pj ∩ H is a maximal subgroup of Pj ; and so by Theorem 6.6 and as Pj is a pj -group

Pj ∩ H Pj .

We can now apply Lemma 4.14(iv). Above we noted that P1 G, and so this lemma gives

P1(Pj ∩ H) P1Pj ;

if j = 1 replace P1 by P2.WealsohaveP2 G, and so we can repeat this argument to obtain P1P2(Pj ∩ H) P1P2Pj . Continuing we have ﬁnally

H P1 ···(Pj ∩ H)···Pk P1 ···Pj ···Pk = G.

This gives (ii). (iii) implies (i) First we prove

P1 ···Pj G and o(P1 ···Pj ) = o(P1) ···o(Pj ), (7.1)

for j = 1,...,k, by induction on j.Forj = 1 this follows by hypothesis, so suppose it is true for j.AsP1 ···Pj G (by the inductive hypothesis) and Pj+1 G by (iii), Lemma 4.14(iii) gives P1 ···Pj+1 G. Secondly

o(P1 ···Pj+1) = o(P1 ···Pj )o(Pj+1)

by Theorem 5.8 as Pj+1 ∩ P1 ···Pj =e and Pj+1 is a Sylow pj+1- subgroup. This shows that (7.1) holds for all j ≤ k, and Condition (ii) of Theorem 7.6 follows if we put j = k. This last conclusion also justiﬁes Con- dition (iii) in Theorem 7.6, and so the result follows. 146 7 Products and Abelian Groups

Example Referring to Appendix C, we note that the group

Q2 × C3 is an example of a nilpotent group (of order 24) satisfying the conditions of the theorem. This group has unique Sylow subgroups isomorphic to Q2 and C3 (of orders 8 and 3, respectively) which are therefore normal, it is a direct product, and all of its maximal subgroups are normal. These subgroups are three copies of C12 (of index 2, and so normal; see Problem 2.19), and the unique Sylow 2-subgroup is Q2 (normal by Theorem 6.10). The group D4 × C3 is another non-Abelian example listed in Appendix C, as an exercise the reader should describe its maximal subgroups.

7.2 Finite Abelian Groups

We now consider finite Abelian groups in more detail. When studying Abelian groups, some authors use an additive notation. We use the multiplicative notation throughout to ‘keep the work in context’. With cyclic groups as factors, many Abelian groups can be constructed using the direct product construction described in the previous section, see Lemma 7.5(ii). Remarkably, the opposite is also true in the finite case, that is, all finite Abelian groups are isomorphic to direct products of cyclic groups. This is not true for infinite groups, for example, the rational numbers Q with addition cannot be expressed as a direct product of cyclic groups; see Web Section 7.5. Several direct products of cyclic groups can be constructed having order n when n is composite, but they may not all be distinct (non-isomorphic). Previously we showed that C6 and C2 × C3 are isomorphic; note that 2 and 3 are coprime. On the other hand, consider the case n = 8. We have three possibilities

C8,C4 × C2 and C2 × C2 × C2, and no two are isomorphic—the first contains an element of order 8 whilst the others do not, and the first two contain elements of order 4 whilst the last is an elementary Abelian 2-group. These two examples are typical. If G is a group of order n, and n is a product of distinct primes then the group is completely determined provided it is Abelian, but if high powers occur in the factorisation of n then many Abelian groups are possible, see the note at the end of this section. The Fundamental Theorem of Finite Abelian Groups characterises these groups completely; it is in two parts. The first part, called the Basis Theorem, describes the essential structure. We give two proofs, the first is relatively short and describes the basic facts, but it does not establish the full picture as the prime factorisation of the group order is involved only indirectly, the second gives more information. The remaining part of the Funda- mental Theorem determines the isomorphism classes, that is, it provides conditions under which two finite Abelian groups are isomorphic, its proof will be given in the problem section (Problem 7.15). 7.2 Finite Abelian Groups 147

A number of proofs of the Basis Theorem use the Euclidean Algorithm (Theo- rem B.2). Our ﬁrst proof relies on the following matrix result which is a consequence of this algorithm. An n×n matrix A is called unimodular if every entry is an integer, and det A = 1 (and so A is non-singular); in this case, A−1 is also unimodular. We have

Lemma 7.10 If r1,...,rn are integers having no positive common factor except 1, then there exists an n × n unimodular matrix with ﬁrst column (r1,...,rn) , where denotes the transpose.

For a proof of this result, see Rose (1999), page 165; the 2 × 2 case is given by the Euclidean Algorithm, and the general case follows by induction. We give now the ﬁrst proof of the Basis Theorem, it relies on the following

Lemma 7.11 If {g1,...,gn} is a generating set for an Abelian group G, and r1,...,rn are coprime integers (that is, they have no common factor larger than 1), then G has a second generating set one of whose elements is

= r1 ··· rn h1 g1 gn .

Proof Let A = (sij ) be one of the unimodular matrices with ﬁrst column (r1,...,rn) given by Lemma 7.10 where

ri = si1 for i = 1,...,n, and let

= s1i ··· sni = hi g1 gn for i 1,...,n. −1 Note that hi ∈ G for all i, as each sij ∈ Z.NowletA = (tij ), then tij ∈ Z for all i and j, and we have si1t1j +···+sintnj equals 1 if i = j, and it equals −1 0ifi = j (as AA = In). Hence +···+ +···+ +···+ = s11t1j s1ntnj ··· sj1t1j sjntnj ··· sn1t1j snntnj gj g1 gj gn s t s t = s11 j1 sn1 1j ··· s1n jn snn nj g1 ...gj ...gn g1 ...gj ...gn = t1j ··· tnj h1 hn ,

as G is Abelian. This shows that the set {h1,...,hn} can be used as a generating set for G, and h1 has the required property.

Using this lemma, we give the ﬁrst proof of

Theorem 7.12 (Basis Theorem for Finite Abelian Groups) Every ﬁnite Abelian group G is isomorphic to a direct product of cyclic groups.

In some cases, the product has only one term, for instance, when G = Cp, see also Theorem 7.7(ii). 148 7 Products and Abelian Groups

Proof By induction on o(G). The result is clearly true for the neutral group; and so, using the inductive hypothesis, we may assume that it holds for all Abelian groups of order less than o(G). As G is ﬁnite, it has a ﬁnite generating set. Let n be the least positive integer such that G has a generating set with n elements; as noted above if n = 1 there is nothing to prove. So we have

every subset of G with at most n − 1 elements fails to generate G.(∗)

Secondly, amongst elements of all n-element generating sets for G choose g1 of least order t, say (note t>1), and let the remaining elements of the generating set containing g1 be g2,...,gn.LetH be the subgroup of G generated by g2,...,gn.By(∗) and the inductive hypothesis, H is a proper subgroup of G, and it is isomorphic to a direct product of cyclic groups using Lemma 7.5(iv). We prove the theorem by showing

G g1×H.

By Theorem 7.4, this will follow if we can prove that

g1∩H =e, (7.2)

because G =g1H by deﬁnition, and all subgroups are normal. Suppose (7.2) is false, that is, a non-neutral element x ∈ G exists satisfying x ∈g1 and x ∈ H . It follows that integers r1,...,rn exist with

= r1 = r2 ··· rn x g1 g2 gn , (7.3) and

where t was deﬁned above, r1 = 0 because x = e, and r1

− = r1/d r2/d ··· rn/d y g1 g2 gn . (7.5)

We know that {g1,...,gn} is a generating set for G and, by Lemma 7.10 and as the GCD of the set of integers {−r1/d, r2/d,...,rn/d} equals 1, the element of G on the right-hand side of (7.5) is a member of a generating set for G with n elements. But by (7.3)

yd = e,

that is, o(y) ≤ d

We come to the second proof of Theorem 7.12, as noted above it is longer but it provides more information. In the ﬁrst part (Lemma 7.13), we show that a ﬁnite Abelian group G is isomorphic to a direct product of pi -subgroups where pi | o(G).

= s1 ··· sk Lemma 7.13 Suppose G is a ﬁnite Abelian group and o(G) p1 pk , where ≥ = = p1,...,pk are distinct primes, and si 1, i 1,...,k. For i 1,...,k, let Hpi denote the subset of all those elements of G whose orders are a power of pi . ≤ (i) Hpi G. ×···× (ii) G Hp1 Hpk .

∈ = j = k Proof (i) If a,b Hpi , o(a) pi , o(b) pi , and if l equals the maximum −1 | l of j and k, then, as G is Abelian, o(a b) pi . Hence (i) follows. (ii) The ﬁrst proof was given in Theorem 7.9—the subgroups Hpi given in (i) are the Sylow pi -subgroups of G, and they are normal because the group is Abelian. For a second more direct proof, we argue as follows using = si Theorem 7.6.Letmi o(G)/pi . The integers m1,...,mk have no common factor larger than one by deﬁnition, so using the Euclidean Algorithm (Theo- rem B.2), integers r1,...,rk exist satisfying

r1m1 +···+rkmk = 1.

Now if x ∈ G, then for each i = 1,...,k, the integer o(xmi ) is a power | mi ∈ of pi (as o(x) o(G)), and so x Hpi . Hence +···+ = m1r1 mkrk = m1 r1 ··· mk rk ∈ ··· x x (x ) (x ) Hp1 Hpk .

As this holds for all x ∈ G, Condition (ii) in Theorem 7.6 is satisﬁed. ∈ ∩ ··· ··· For the third condition, suppose y Hpi Hp1 Hpi−1 Hpi+1 Hpk .So there exist integers si satisfying

= si = s1 ··· si−1 si+1 ··· sk o(y) pi p1 pi−1 pi+1 pk .

But this is only possible if s1 = ··· = sk = 0(theprimespi are distinct), in which case y = e and so Condition (iii) of Theorem 7.6 is also satisﬁed. Condition (i) is automatically satisﬁed, so the proof is complete.

Continuing the second proof of Theorem 7.12, by Lemma 7.13 we now need to prove the result in the case when G is a p-group, and so we can make use of the work on these groups given in Section 6.1.

Proof of Theorem 7.12 in the case when G is a ﬁnite Abelian p-group, (and so by Theorem 6.3, when o(G) is a power of p) We use induction on o(G), there is nothing to prove if o(G) ≤ p. Let H be a maximal subgroup of G. By Theorem 6.6, o(G/H) = p and, by the inductive hypothesis, we can express H as a direct product of j,say, 150 7 Products and Abelian Groups

r cyclic p-groups Hi where o(Hi) = p i , i = 1,...,j; that is, by the inductive hypothesis we have

H H1 ×···×Hj and r1 ≥···≥rj ≥ 1. (7.6)

This holds by rearranging the Hi ; see Lemma 7.5(iv). Choose a ∈ G\H and then ap ∈ H , there are p cosets at H , one for each t in the range 0,...,p− 1. Hence we can choose hi ∈ Hi to satisfy

p a = h1 ···hj . (7.7)

We may assume that each hi is either a generator of Hi , or equals e. For by = p Problem 6.1(ii), a non-generator of Hi is a pth power, and if hi bi for some bi ∈ Hi , then using (7.6) and as G is Abelian we have −1 p = p −p = p −1 = ··· ··· abi a bi a hi h1 hi−1hi+1 hj . −1 ∈ But abi H by deﬁnition of a and bi ; and so this contradiction establishes our assumption. Now if ap = e, then G is isomorphic to a direct product of a, H1,...,Hj−1 and Hj , and the theorem follows in this case. Hence we

may assume that hi = e for some i satisfying 1 ≤ i ≤ j.Leti equal the p smallest such i,sohi = e and a = hi ···hj by (7.7). p r Now by (7.7)again,o(a ) = p i (order is LCM of o(hi ),...,o(hj )), r +1 hence o(a) = p i .LetJ = H1 ×···×Hi −1 ×e×Hi +1 ×···×Hj . Clearly, o(J) = o(H)/pri , and the theorem will follow if we can show that

a∩J =e, (7.8)

for then aJ is a direct product with order equal to o(G). Suppose (7.8) is false, and so at ∈ J for some t.Nowp | t by deﬁnition. So if t = sp for some s satisfying 0 ≤ s

by deﬁnition of s. This shows that the unique (we have a direct product) rep- t resentation (7.9)ofa given in H = H1 ×···×Hj has a non-neutral element in its i th place. But this cannot belong to J because J has e here. This establishes (7.8), and the main result follows.

The second proof of Theorem 7.12 is now complete. For, by Lemma 7.13, a ﬁnite Abelian group G is isomorphic to a direct product of pi -subgroups where pi | o(G), and each of these is itself a direct product of cyclic pi -subgroups. Both of these proofs rely on an argument involving induction. We apply the inductive hypothesis to G with one of its smallest terms H , say, removed, to obtain a direct product K, and then show that we can form the direct product of H and K. Other proofs work by removing the largest factor. 7.3 Semi-direct Products 151

Example Consider the case n = 72. We have 72 = 23 · 32, and each Abelian group of order 72 is a direct product of a group H of order 8 and a group J of order 9 (Lemma 7.13). There are three possibilities for H : C8,C4 × C2 and C2 × C2 × C2, and two for J : C9 and C3 × C3. Hence there are six (isomorphism classes of) Abelian groups of order 72:

C8 × C9,C8 × C3 × C3,C4 × C2 × C9,C4 × C2 × C3 × C3,

C2 × C2 × C2 × C9, and C2 × C2 × C2 × C3 × C3.

This list is complete; for example, C18 × C4 is included because C18 C2 × C9, and so C18 × C4 C4 × C2 × C9. Note also that the groups Tn,forn = 1, 2,..., introduced in Section 2.2 can now be deﬁned by Tn = C2 ×···×C2 with n copies of C2. In general, if n>1 and part(n) denotes the number of partitions of n (that is, the number of ways of writing n as a sum of equal or smaller positive integers; for example, 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1, and so part(4) = 5), then there are part(n) distinct (non-isomorphic) Abelian groups of order pn. The integer part(n) increases exponentially with n; for instance, there are 42 (isomorphism classes of) Abelian groups of order 1024, and 627 of order 220. For further details on partitions, see Andrews (1976).

7.3 Semi-direct Products

As noted earlier, given two groups H and J , a number of new groups can be constructed on the underlying set H × J . One is the direct product, but in many cases others are possible. The new group is, in a sense to be made precise, either an extension of H by J ,orofJ by H ; see Chapter 9. Contrary to our discussion of direct products given in Section 7.1, we begin our work with the internal semi-direct product, this will lead to the ‘right’ deﬁnition for the external product. We take the same underlying set as in the direct case and deﬁne a different operation, and we begin with (cf. Theorem 7.4)

Deﬁnition 7.14 Let G be a group with a subgroup A and a normal subgroup K which satisfy G = AK and A ∩ K =e. (7.10) In this case, G is called an (internal) semi-direct product of K by A, and we write

G A K.

n 2 n−1 For example, the dihedral group Dn =a,b | a = b = e,bab = a can be represented as a semi-direct product if we take b for A and a for K, and then Dn b a. 152 7 Products and Abelian Groups

Notes (a) In symbols we always place the normal subgroup K second next to the triangular part of the sign. The subgroup A is called a complement of K in G, and G is sometimes called a split extension of K by A. Some authors put the normal subgroup K ﬁrst and write K A.TheATLAS (1985)usesK : A for our A K (and K.A for a general extension of K by A). (b) If G is the direct product of A and K, then clearly it is also a semi-direct product of K by A, and of A by K. (c) G can only be simple if either A or K is the neutral subgroup, also we shall see below (Lemma 7.15) that G is not Abelian when the product is not direct even if both A and K are Abelian. (d) It is important to note that given A and K, the group G is not uniquely determined; for example, both D3 and C6 are semi-direct products of C3 by C2.Note also that some groups can be expressed as a semi-direct product in several different ways; see page 183, for example. (e) There are non-simple groups that cannot be expressed as semi-direct products, for example, the quaternion group Q2. This group has a number of normal subgroups Ki with corresponding complements (subgroups) Ai ,but(7.10) fails in each case; see Problem 7.17. Direct and semi-direct products do provide a number of interesting groups, but many more constructions need to be considered if all groups are to be described. (f) A number of authors have considered the question: when can a group G be expressed as a product of two of its proper subgroups (with no restriction on these subgroups except that they are both proper), that is, when is G factorisable; see, for example, Scott (1964), Chapter 13. Clearly Cp is not factorisable, but Scott gives other instances—one is the simple group L2(13), see Section 12.2.Healsogives some positive results which we shall discuss in Section 11.2.

Note (d) above suggests that something more is needed. The following lemma provides this missing link.

Lemma 7.15 Suppose G = A K. (i) If G is ﬁnite, o(G) = o(A)o(K). (ii) If g ∈ G, then g can be uniquely represented by g = ak where a ∈ A and k ∈ K. (iii) If gi = aiki where ai ∈ A and ki ∈ K, for i = 1, 2, then g1g2 has the unique = = ∈ = −1 ∈ representation as g1g2 ak, where a a1a2 A and k a2 k1a2k2 K. (iv) There is a homomorphism γ : A → Aut K with the property that the product g1g2 given in (iii) can be deﬁned by

g1g2 = (a1k1)(a2k2) = (a1a2)(k1(a2γ ))k2.

(v) The homomorphism γ given in (iv) is the trivial map (see page 69) if and only if the product is direct. (vi) If the homomorphism γ given in (iv) is not the trivial map, then G is not Abelian. 7.3 Semi-direct Products 153

Proof For gi ∈ G we write gi = aiki throughout where ai ∈ A and ki ∈ K; see Definition 7.14. (i) By the Second Isomorphism Theorem (Theorem 4.15) and (7.10)we have A A/A∩K AK/K = G/K, and so the result follows by Lagrange’s Theorem (Theorem 2.27). (ii) Suppose g = a1k1 = a2k2, ai ∈ A and ki ∈ K,fori = 1, 2. Then −1 = −1 = = = a2 a1 k2k1 e by (7.10), that is a1 a2 and k1 k2. = = −1 ∈ (iii) We have g1g2 a1k1a2k2 a1a2a2 k1a2k2, where a1a2 A and, as −1 ∈ −1 ∈ K G, a2 k1a2 K, and so a2 k1a2k2 K. Uniqueness follows from (ii). (iv) Define the map γ : A → Aut K using conjugation as follows. If a ∈ A and k ∈ K,let k(aγ) = a−1ka. By definition, aγ is an automorphism of K for each a ∈ A, and so k(aγ) ∈ K also for each a ∈ A.Nowγ is a homomorphism because a1γ ◦ a2γ = −1 = −1 −1 (a1a2)γ , this follows because (a1a2) k(a1a2) a2 (a1 ka1)a2. (v) First, γ is the trivial map on A if, and only if, aγ equals the identity map (automorphism) on K for all a ∈ A, that is k(aγ) = k for all a ∈ A and k ∈ K. If the product is direct, then ak = ka for all a ∈ A and k ∈ K, and so γ is the trivial map. Conversely, if γ is the trivial map, then for all k we have k(aγ) = k,butk(aγ) = a−1ka, and so ak = ka for all a ∈ A and k ∈ K.This shows that A G, hence the product is direct by Theorem 7.4. (vi) By (iv), if γ : A → Aut K is not trivial, there is an a ∈ A such that aγ is not the identity map, hence there exist a ∈ A and k ∈ K such that k(aγ) = k. But by (iv) k(aγ) = a−1ka, that is, a and k do not commute.

These results suggest that to deﬁne an external semi-direct product of K by A we need to consider the homomorphism γ ; in many cases different γ will give rise to different groups even if both A and K are ﬁxed.

Deﬁnition 7.16 Given groups A and K, and a homomorphism γ : A → Aut K,the (external) semi-direct product A γ K of K by A relative to γ is the group with underlying set A × K, and operation (a1,k1)(a2,k2) = a1a2, k1(a2γ) k2 , (7.11) where ai ∈ A and ki ∈ K,fori = 1, 2.

Note that a2γ is an automorphism of K for each a2 ∈ A, and so k1(a2γ)∈ K, hence we also have (k1(a2γ ))k2 ∈ K when k1,k2 ∈ K. Before stating the next result (Theorem 7.17) we consider the following

Example Let A =aC2 and K =kC3, see (d) opposite. By Theorem 4.23, Aut C3 C2, that is C3 has two automorphisms η1 and η2 as follows: 154 7 Products and Abelian Groups

t u u (i) η1 is the identity map which satisﬁes a η1 : k → k for all k ∈ K and t,u∈ Z. u u (ii) η2 is the map which satisﬁes, for all integers u, eη2 : k → k and aη2 : u → 2u 2 k k , and so η2 is the identity map η1 on K. Combining the two statements in (ii) we obtain: for t = 0or1,andforallu

t u (t+1)u a η2 : k → k . (7.12)

We can deﬁne two groups using (i) or (ii), respectively. In the ﬁrst case, s t s k (a η1)=k for all s and t, which gives for 0 ≤ r, t ≤ 1, 0 ≤ s,u ≤ 2, r s t u r+t s t u r+t s+u a ,k a ,k = a , k a η1 k = a ,k , the direct product of A and K isomorphic to C6. For the second group, we use (7.12). We have, for r, s, t and u in the same ranges as those listed above, r s t u r+t s t u r+t (t+1)s+u a ,k a ,k = a , k a η2 k = a ,k .

The reader should check that this gives a group isomorphic to the dihedral group D3 where the usual generators a and b are replaced by k and a, respectively.

The next result shows that the semi-direct product deﬁnes a group.

Theorem 7.17 (i) The set with operation in Definition 7.16 forms a group. (ii) The map γ given in Definition 7.16 is defined by conjugation.

To make this proof easier to follow we have reintroduced the symbol to denote the group operation in K, that is, (7.11) now reads (a1,k1)(a2,k2) = a1a2, k1(a2γ) k2 .

Proof (i) The operation is well-deﬁned by deﬁnition. The neutral element is (e, e) for (a, k)(e, e) = ae, k(eγ) e = (a, k)

as eγ is the identity map. The inverse of (a, k) is (a−1,k−1(a−1γ))for (a, k) a−1,k−1 a−1γ = aa−1, k a−1γ k−1 a−1γ = e, k k−1 a−1γ = (e, e)

because a−1γ is a homomorphism (automorphism) which maps e to e.For associativity, we proceed as follows: The square brackets in the expressions 7.3 Semi-direct Products 155

below are not strictly necessary (they enclose products in K) but are inserted to aid clarity. We have, where ai ∈ A and ki ∈ K for i = 1, 2, 3, (a1,k1)(a2,k2) (a3,k3) = a1a2, k1(a2γ) k2 (a3,k3) = a1a2a3, k1(a2γ) k2 (a3γ) k3 ,

and (a1,k1) (a2,k2)(a3,k3) = (a1,k1) a2a3, k2(a3γ) k3 = a1a2a3, k1 (a2a3)γ k2(a3γ) k3 = a1a2a3, k1 (a2a3)γ k2(a3γ) k3 ,

by associativity in K. The expressions on the right-hand sides of these equations are equal because k1 (a2a3)γ k2(a3γ) = k1(a2γ ◦ a3γ) k2(a3γ) as γ is a homomorphism = k1(a2γ) (a3γ) k2(a3γ) by deﬁnition of composition = k1(a2γ) k2 (a3γ) as a3γ is an automorphism.

This proves (i). (ii) The set {(a, e) : a ∈ A} forms an isomorphic copy of A in G, similarly the set {(e, k) : k ∈ K} forms an isomorphic copy of K in G. Now using the equations above we have (a, e)−1 = a−1,e(a−1γ) = a−1,e ,

because a−1γ is a homomorphism, and (e, k)(a, e) = ea, k(aγ) e = a,k(aγ) .

Hence (a, e)−1(e, k)(a, e) = a−1,e a, k(aγ) = a−1a, e(aγ ) k(aγ) = e,k(aγ )

for all a ∈ A and k ∈ K, and as e(aγ ) = e. Therefore, the homomorphism γ is given by conjugation of an element in K by a element in A.

As an exercise the reader should show directly (using the same methods as above) that (e, e)(a, k) = (a, k) and (a−1,k−1(a−1γ ))(a, k) = (e, e); note these equations also follow from Theorems 7.17 and 2.5. 156 7 Products and Abelian Groups

Wreath Product

Before providing some examples we give a brief introduction to a particular type of semi-direct product called a wreath product, a number of groups with special properties can be deﬁned using it. Suppose G and H are groups, X ={1, 2,...,n}, H acts on X, and G∗ is the direct product of n copies of G. We deﬁne an action of H on G∗ by \ = ∈ (g1,...,gn) h g1\h−1 ,...,gn\h−1 for h H.

It is easy to see that the action axioms (5.1) hold, and \ \ = \ (g1,...,gn) h g1,...,gn h g1g1,...,gngn h, gives a homomorphism. The wreath product of G by H is deﬁned as the semi- ∗ direct product H φ G where φ is the homomorphism given above. The product is denoted by either G wr H or G H . We have (a) the product is associative, and n (b) o(G wr H)= o(H) · o(G) . An example is: C2 wr C2 D4. We mention two applications.

(a) If G is soluble with derived length n (page 234), then G wr C2 has derived length n + 1, hence there exist soluble groups with arbitrarily high derived length; for a proof, see the ﬁrst reference given below. (b) The Sylow subgroups of the symmetric groups Spm are isomorphic to iterated (m − 1 times) wreath products of the cyclic group Cp, and the Sylow subgroups of general symmetric groups can be constructed using direct products of these iterated wreath products; this is proved in the last reference below. For more details and proofs, the reader should see Rose (1978), page 219 ff.Rotman (1994), page 173 ff, Suzuki (1982), page 268 ff, and Cameron (1999) should also be consulted for (b).

Groups of Order 12

As an application of the theorems above, we consider the ﬁve (isomorphism classes of) groups of order 12; in fact, all of these groups can be treated as semi-direct products. By Theorem 7.12, there are two (isomorphism classes of) Abelian groups of order 12:

C12 C4 × C3 and C6 × C2 C3 × C2 × C2; these are direct products (and so are special cases of semi-direct products). For non- Abelian groups, we have the following preliminary 7.3 Semi-direct Products 157

Lemma 7.18 If G is a non-Abelian group of order 12, then G is either (i) isomorphic to A4, or (ii) it contains a normal cyclic subgroup of order 6.

Proof Let P be a Sylow 3-subgroup of G.Aso(P ) = 3, P is cyclic, and so we can assume that it has the form a where a3 = e.Wealsohave[G : P ]= 4, and so by Theorem 5.15, there is an injective homomorphism

−1 θ : G h Ph→ S4. h∈G

−1 Now h∈G h Phequals e or P (as o(P ) is prime). If the ﬁrst case applies then θ is an injective homomorphism of G into S4, and so G is isomorphic to a normal (as the index is 2) subgroup of S4. Problem 3.3 now shows that G A4. Note that A4 has no elements of order 6. −1 = For the second case, we have h∈G h Ph P , and so P G.Thisim- plies that G has only one (Sylow) subgroup of order 3 (Sylow 3), and so G has only two elements of order 3: a and a2. This further implies that a can only have one or two conjugates in G as the order of a conjugate of a equals the order of a. Theorem 5.19 now gives [G : CG(a)]=1or2, where CG(a) denotes the centraliser of a in G. Hence o(CG(a)) = 12 or 6. By Cauchy’s Theorem (Theorem 6.2), in either case CG(a) contains an element b, say, of order 2, and this element commutes with a. Therefore, ab ∈ G, o(ab) = 6, and so ab is a normal (as its index in G is 2) subgroup of G of order 6.

Continuing the argument given in the second part of the above proof, let ab = c, then o(c) = 6 and c G. Suppose d ∈ G\c, then

− d 1cd = ct for some t = 0,...,5.

If t = 0 then c = e,ift = 1 then G is Abelian, if t = 2 then e = (d−1cd)3 = d−1c3d which implies c3 = e, and if t = 3 or 4 a similar argument shows that c2 = e or c3 = e. All of these contradict the properties of c given above, and so t = 5isthe only possibility. Secondly, as the elements cr and cr d,forr = 0,...,5, are all distinct in G (the reader should check this), and o(G) = 12, we have d2 ∈c.Ifd2 = c or c5 then o(d) = 12 and G is cyclic, hence we may exclude these possibilities. If d2 = e then

6 2 5 G D6 = a,b | a = b = e,bab = a . 2 = 2 = 2 2 = 2 2 = Secondly, if d c then, if we replace d by d1 c d,wehaved1 (c d) e −1 = −1 7 = 5 and d1 cd1 d c d c , and so again G D6 now with generators c and d1. 2 4 Thirdly, if d = c , arguing similarly we again have G D6. The ﬁnal possibility, 2 3 that is, d = c , gives the dicyclic group Q3 of order 12 (Section 3.4) where

6 2 3 −1 5 Q3 = c,d | c = e,d = c and d cd = c . 158 7 Products and Abelian Groups

To recap, we have shown that there are at most three (isomorphism classes of) non-Abelian subgroups of order 12: A4,D6 and Q3. As an exercise the reader should write down the Sylow subgroups of these groups (Problem 7.23). We show now that each of these groups can be deﬁned using the semi-direct product construction, that is,

A4 C3 T2,D6 C2 C6, and Q3 C4 C3.

Case 1: A4.

The group A4 (treated as a permutation group on {1, 2, 3, 4}) has a normal subgroup K1 of order 4 containing the three products of two 2-cycles and e, and so it is isomorphic to the 4-group T2. Taking B1 =(1, 2, 3),wehave

B1 ≤ A4,K1 A4,B1 ∩ K1 =e, and A4 = B1K1.

Reader, why does B1K1 have twelve elements? Hence A4 B1 K1, and we need to construct the homomorphism γ1 : B1 → Aut K1.WehaveAutK1 S3 (see Section 4.4, note that K1 is an elementary Abelian 2-group), and so the only non-trivial homomorphism from B1 (of order 3) to Aut K1 (of order 6) is an injection of B1 onto the (unique) subgroup of S3 generated by its 3-cycles and isomorphic to C3. For example, using (7.11) on page 153 with a1 = (1, 2, 3), k1 = (1, 2)(3, 4), k2 = (1, 3)(2, 4) and a2 ranging over the elements e,(1, 2, 3) and (1, 3, 2) of B1 (they are underlined below) we have

(1, 2, 3)(1, 2)(3, 4) · e(1, 3)(2, 4) = (1, 2, 3) · (1, 2)(3, 4)(1, 3)(2, 4),

(1, 2, 3)(1, 2)(3, 4) · (1, 2, 3)(1, 3)(2, 4) = (1, 2, 3)2 · (1, 4)(2, 3)(1, 3)(2, 4),

(1, 2, 3)(1, 2)(3, 4) · (1, 2, 3)2(1, 3)(2, 4) = e · (1, 3)(2, 4)(1, 3)(2, 4).

This illustrates the fact that we can deﬁne the image of (1, 2)(3, 4) under γ1 using (7.11) by:

eγ1 : (1, 2)(3, 4) → (1, 2)(3, 4),

(1, 2, 3)γ1 : (1, 2)(3, 4) → (1, 4)(2, 3), and

(1, 3, 2)γ1 : (1, 2)(3, 4) → (1, 3)(2, 4), with similar expressions for the images of e,(1, 3)(2, 4) and (1, 4)(2, 3) which the reader should write out. See Problem 3.10 for further properties on A4.

Case 2: D6.

By Lemma 7.18 and the discussion given below its proof, the group D6 contains a normal cyclic subgroup (of order 6) K2 c, and an element d of order 2. If we let B2 =d,wehaveD6 = B2K2 and B2 ∩ K2 =e, as required for a semi-direct 7.4 Problems 159 product. As above we need to construct the homomorphism γ2 : B2 → Aut K2.By Theorem 4.23, and Problems 4.19 and 7.10,AutK2 C2, and the only non-identity 5 automorphism of K2 maps c to c ; that is, eγ2 is the identity automorphism on K2, s 5s and dγ2 is the automorphism that maps c to c . Hence we have for r = 0 or 1, and s,u ∈{0,...,5}, r s 0 u r+0 s u r s+u d c · d c = d c (eγ2) c = d c , r s u r+1 s u r+1 5s+u d c · dc = d c (dγ2) c = d c , which gives a representation of D6. Note that D6 can also be represented by semi- direct product T2 C3, this is a consequence of the fact that D6 is isomorphic to C2 × D3.

Case 3: Q3. 2 The group Q3 contains a normal cyclic subgroup K3 =c of order 3, and a second non-normal cyclic subgroup B3 =d of order 4. Clearly, we have B3 ∩ K3 =e and B3K3 = Q3, and so Q3 is a semi-direct product of K3 by B3. By Theorem 4.23,AutK3 C2, and the only non-trivial homomorphism γ3 of B3 =d to Aut K3 associates the even powers of d with the identity automorphism, and the odd powers of d with the automorphism that maps cs to c2s . This is mirrored in Q3 for we have in this group r s 2t u r+2t s 2t u r+2t s+u d c · d c = d c d γ3 c = d c , r s 2t+1 u r+2t+1 s 2t+1 u r+2t+1 2s+u d c · d c = d c d γ3 c = d c for 0 ≤ r<4, 0 ≤ t<2 and 0 ≤ s,u < 3. The reader should check that this construction gives a group isomorphic to the dicyclic group Q3 deﬁned on page 59.It can also be represented as a metacyclic group, see Theorem 6.18.

7.4 Problems

Problem 7.1 (i)GiveproofsofLemma7.5 and Theorem 7.6. (ii) Show that if G is a product of its subgroups H1,...,Hn, where each Hi G and (o(Hi), o(Hj )) = 1ifi = j, then G H1 ×···×Hn.

Problem 7.2 (i) If G, H, J and K are groups, show that [G × H,J × K] [G, J ]×[H,K], Problem 2.17. (ii) If H G and J K, show that H × J G × K, and

(G × K)/(H × J) G/H × K/J.

(iii) Prove that if J,K G and G = JK, then J/(J ∩ K) × K/(J ∩ K) G/(J ∩ K). 160 7 Products and Abelian Groups

Problem 7.3 Suppose G = H × J . (i) Show that H J if and only if a subgroup D of G can be found which satisﬁes G = HD= JD and H ∩ D = J ∩ D =e. (ii) If H ≤ L ≤ G show that L H × (J ∩ L), see Lemma 7.8.

Problem 7.4 Suppose G = H1 ×···×Hn. Prove the following results.

(i) Z(G) Z(H1) ×···×Z(Hn). ×···× (ii) G H1 Hn. (iii) If K is a perfect normal subgroup of G (Problem 4.8), then K (H1 ∩ K) ×···×(Hn ∩ K) (cf. Lemma 7.8).

Problem 7.5 Let G be a finite Abelian group, and let ex(G) denote its exponent. (i) Show that there exists g ∈ G with the property ex(G) | o(g). (ii) Using (i) show that if ex(G) = o(G), then G is cyclic. (iii) Use (ii) and the fact that in a field a polynomial of degree r has at most r roots (Theorem B.13) to show that the multiplicative group of a finite field is cyclic.

Problem 7.6 (i) Suppose J ≤ G × H , show that J is Abelian, or J intersects one of the factors G or H non-neutrally. (ii) By considering an Abelian group G with order p5 and three of its subgroups 2 H1,H2 and H3, each with order p , show that the following properties can all hold for a suitably chosen group G.

(a) G = H1H2H3, (b) H1H2, H2H3 and H3H1 are all proper subgroups of G, (c) H1 ∩ H2 = H2 ∩ H3 = H3 ∩ H1 =e, and (d) G is isomorphic to a proper subgroup of H1 × H2 × H3.

Problem 7.7 If G is Abelian, o(G) = n, and m | n, show that G has a subgroup of order m. First, reduce to the case when n is a prime power.

Problem 7.8 (i) How many Abelian groups (up to isomorphism) are there of order 385 or 432? (ii) You are given: there are 14 (isomorphism classes of) groups of order 81. Using the Sylow and direct product theories count the number of (isomorphism classes of) groups of order (a) 891, and (b) 405.

Problem 7.9 Suppose G is a ﬁnite group, and all of its maximal subgroups are both simple and normal. Show that G is Abelian, and o(G) = 1,p,p2 or pq where p and q are prime.

Problem 7.10 Show that if G and H are ﬁnite groups and (o(G), o(H )) = 1, then Aut(G × H) Aut G × Aut H . 7.4 Problems 161

Problem 7.11 Suppose G is a ﬁnite Abelian group with proper subgroup H ;see also Problem 7.5. (i) Choose g ∈ G with the largest possible order n. Prove that hn = e for all h ∈ G. (ii) Show that the result given in (i) is false in general for non-Abelian groups, that is, the exponent (Deﬁnition 2.19) can be larger than n. (iii) Suppose J is maximal subject to the conditions J ≤ G and H ∩ J =e.If there exists g ∈ G such that gp ∈ J for some prime p, show that G = HJ. (Hint. Consider the equation h = jgr where h ∈ H , j ∈ J and r ∈ Z.) (iv) With J as in (iii), show that G H × J if and only if for all primes p, and p for all g ∈ G, h ∈ H and j ∈ J with g = hj , there exists h1 ∈ H satisfying = p h h1 . (v) Finally, with g as in (i), show that g is a direct factor of G, that is, a subgroup J ≤ G exists with the property G g×J . (Hint. Use (i) and (iii) and consider the cases p | n and p n separately.)

Problem 7.12 A group G is called characteristically simple CS if its only characteristic subgroups are e and G itself. For the definition of characteristic, see Problem 4.22.LetG be a finite Abelian group. Show that G is CS if and only if it is elementary. A definition of ‘elementary’ is given in Problem 4.18, the note at the bottom of page 235 is also relevant. One method is as follows. First, suppose G is CS and p | o(G).LetH ={g ∈ G : gp = e}. Show that H is CS, and so deduce that H = G. For the converse, use the definition of an automorphism, and Problem 4.18.

Problem 7.13 (Hamiltonian Groups) A ﬁnite non-Abelian group is called Hamilto- nian if all of its proper subgroups are normal; such groups are named after Hamilton, the discoverer of the quaternion groups, see page 119. R. Dedekind (1831Ð1916) proved that G is Hamiltonian if and only if it can be expressed as a direct product:

G Q2 × T × O, where Q2 is the quaternion group (of order 8), T is an elementary Abelian 2-group (so T Tn for some n, see Problems 4.18), and O is an arbitrary ﬁnite Abelian group of odd order. You are asked to show that if G has this form, then it is Hamil- tonian. For the converse, see Robinson (1982), page 138. (Hint. Apply Lemma 7.8, consider the cases when the subgroup does, or does not, intersect Q2, and use Prob- lem 2.16(iii).)

Problem 7.14 Let G be a ﬁnite Abelian group, and let mi ∈ Z for i = 1,...,k. ∈ m1 ··· mk = Elements a1,...,ak G are said to be independent if a1 ak e implies mi = ≤ ≤ ai e, for all i satisfying 1 i k.

(i) Show that the elements a1,...,ak are independent if and only if a1,...,ak a1×···×ak. (ii) Deduce G Z/pZ ×···×Z/pZ,ifG has exponent p. Groups of this type are called elementary Abelian, see Problem 4.18. 162 7 Products and Abelian Groups

Problem 7.15 (Isomorphism Theorem for Finite Abelian Groups) In this problem, assume that all groups are ﬁnite Abelian p-groups unless stated otherwise. For a ﬁnite Abelian group G,letd(G) denote the minimum number of generators of G. (i) Show that if G and H are elementary, then d(G× H)= d(G)+ d(H). (ii) Suppose G has the decomposition G C1 ×···×Ck where each Ci is cyclic. n pn pn+1 Show that the number of factors Ci with order larger than p is d(G /G ) where as usual Gm ={gm : g ∈ G}. (iii) For n ≥ 0, let pn pn+1 pn+1 pn+2 up(n, G) = d G /G − d G /G .

Show that up(n, G) depends on G but is independent of the particular decomposition, this gives the number of cyclic factor groups of order pn+1 in all decompositions of G. (iv) Deduce G and H are isomorphic if and only if up(n, G) = up(n, H ) for all n ≥ 0. (v) Give a condition to determine whether two general ﬁnite Abelian groups are isomorphic.

Problem 7.16 Suppose G = AK and K ≤ J ≤ G. Show that J = (A∩J)K.

Problem 7.17 Which of the following groups can be represented as semi-direct products: (i) C15, (ii) S4, (iii) Q2? Give reasons.

(iv) Show that GL2(Q) can be represented as a semi-direct product of SL2(Q) by Q∗. (v) Give an example of a group which can be represented in two distinct ways as a semi-direct product.

Problem 7.18 Let H and J be groups where H is cyclic, and let φ and ψ be injective homomorphisms from H to Aut J subject to the condition Hφ = Hψ. Show that H φ J H ψ J . Applications of this result are given in Problems 7.21 and 7.22.

Problem 7.19 (i) Suppose G is a group and SG is the full permutation group de- ﬁned on the elements of G. Further, for a ∈ G,letξa : G → G be the map given ∗ by gξa = ga (Cayley’s Theorem (Theorem 4.7)), and let G ={ξa : a ∈ G}.Note ∗ that G ≤ SG.Theholomorph of G, denoted by Hol(G), is deﬁned as the group generated by G∗ and Aut G. Show that Hol(G) = Aut G G∗.(Hint.Ifφ ∈ Aut G, −1 then φ ξaφ = ξaφ.) (ii) Find the holomorphs of the cyclic groups Cn when n = 1,...,6; for C5 use Problem 6.19.

Problem 7.20 Investigate the semi-direct products of the cyclic group C4 with itself. You should determine how many (isomorphism classes) there are, describe them, and list their elements, subgroups and normal subgroups. 7.4 Problems 163

Problem 7.21 Let G be a group of order pq where p and q are primes and p

Problem 7.22 List the subgroups of the groups of order 12 discussed at the end of this chapter.

Problem 7.23 Suppose G is a non-Abelian group of order p3 which contains an element of order p2; see Problem 6.5. Using the method suggested below show that G Cp Cp2 , and that all such semi-direct products are isomorphic. The method is as follows. (a) Choose a,b ∈ G with o(b) = p2 and a ∈ b.Ifap = e we have the desired semi-direct product, why? (b) Suppose ap = e. Show that o(Z(G)) = p, and so Z(G) < b. Further show p tp that G/Z(G) Cp × Cp (use Problem 4.16(ii)), and so deduce that a = b for some t with 1 ≤ t

Problem 7.24 (Project—A Group of Order 84) Suppose G is a group with o(G) = 84 having the maximum possible number (28) of Sylow 3-subgroups. You are asked to determine G; there is only one isomorphism class; see also Scott (1964), page 218. (i) Using the Burnside’s Normal Complement Theorem (Theorem 6.17) show that G contains a normal subgroup J of order 28. (ii) Show that G has a normal Sylow 7-subgroup K, and that CG(K) = J using the N/C-theorem (Theorem 5.26) and (i). (iii) Show that G has a unique Sylow 2-subgroup H which is isomorphic to C2 × C2. (Hint. Use (i).) (iv) Using (i), (ii) and (iii) deduce J C2 × C2 × C7. (v) Next show that G can be treated as a semi-direct product involving J . Ideas related to holomorphs will be useful here, see Problem 7.19. (vi) Finally show that G has a presentation in the form

G a,b,c,d | a2 = b2 = c7 = d3 = e,ab = ba,bc = cb,ca = ac,

d2ad = b,d2bd = ab,d2cd = c2 . Chapter 8 Groups of Order 24 Three Examples

In Chapter 1, we commented that both theorems and examples are important when studying groups, and so as an interlude before we introduce our next major theoretical topics we shall consider three groups of order 24 in detail to illustrate the material from the previous six chapters and motivate the work on series, simple groups, and (on the web) representation and character theory to come. ‘Getting to know’ the structure of these groups ‘in full’ will, we are sure, help the reader in his (her) general understanding of the theory as a whole. Because our work on character theory is being given in the Web Sections, we shall give the character tables for the three groups discussed in this chapter in Web Section 14.1. Some facts about the (isomorphism classes of the) remaining twelve groups of order 24 are tabulated in the Appendix C, these include details about their subgroups (including special subgroups), factor groups, automorphisms, and some more specialised properties. The reader should use this chapter to experiment with various hypotheses. For example, what is the relation between the centre and the derived subgroup of a group? Whilst reading this book similar questions will arise, many of which can be an- swered by considering the properties of the groups discussed here. In this chapter a number of facts are not fully justiﬁed, the reader is asked to ﬁll in the details.

8.1 Symmetric Group S4

For our ﬁrst example, we consider the symmetric group S4, and use the results concerning these groups proved in Chapter 3. Our main approach will be via permutations of the set X ={1, 2, 3, 4} although, as we shall show later, the group has several distinct representations; see page 170.Wehaveo(S4) = 24 because there are 4! ways of ordering the set X. Also in a symmetric group, elements are conjugate if and only if they have the same cyclic structure (Theorem 3.6). In the table overleaf which lists the elements of S4, the conjugacy classes are given by the rows or extended rows. H.E. Rose, A Course on Finite Groups, 165 Universitext, DOI 10.1007/978-1-84882-889-6_8, © Springer-Verlag London Limited 2009 166 8 Groups of Order 24—Three Examples

Number Cyclic List of elements of elements type

11-cyclee 62-cycle(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4) 83-cycle(1, 2, 3), (1, 3, 2), (1, 2, 4), (1, 4, 2), (1, 3, 4), (1, 4, 3), (2, 3, 4), (2, 4, 3) 64-cycle(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2) 32-cycle× 2-cycle (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)

Subgroups of S4

The group S4 has thirty subgroups in eleven conjugacy classes. First, we list the cyclic subgroups. Each element of order 2 generates a cyclic subgroup of order 2, hence

S4 has nine subgroups isomorphic to C2 in two conjugacy classes: (1, 2), (1, 3), (1, 4), (2, 3), (2, 4), and (3, 4); and (1, 2)(3, 4), (1, 3)(2, 4) and (1, 4)(2, 3). There are also eight elements of order 3, and so (Problem 2.5(iii))

S4 has four subgroups isomorphic to C3 in one conjugacy class: (1, 2, 3), (1, 2, 4), (1, 3, 4) and (2, 3, 4); see the Sylow entry below. The six elements of order 4 give

S4 has three subgroups isomorphic to C4 in one conjugacy class: (1, 2, 3, 4), (1, 2, 4, 3) and (1, 3, 2, 4).

There are no other cyclic subgroups because S4 has no elements of order larger than 4.

Sylow Subgroups

As with all groups of order 24, S4 has Sylow subgroup(s) of order 3 and 8. The second item above lists the four Sylow 3-subgroups. (Note that in this case the Sylow theory states: n3 ≡ 1 (mod 3) and n3 | 8, and so n3 = 1or4.)AsS4 has more than one Sylow 3-subgroup, none is normal, but they form a single conjugacy class by Sylow 2. For the second case, we have n2 ≡ 1 (mod 2) and n2 | 3. Therefore, n2 = 1or3, and S4 has one or three Sylow 2-subgroup(s) of order 8. First, we need to determine the type, see Theorem 6.10 (Sylow 4) and the example on page 125. This subgroup cannot be C8 because S4 has no elements of order 8. Also it cannot be C2 × C2 × C2 8.1 Symmetric Group S4 167 because such a subgroup would contain seven of the nine elements of order 2 in S4, and these would generate S4; see Problem 3.1. Further, it cannot be Q2 because such a subgroup would contain the six elements of order 4, and a single element of order 2; but this is impossible because the conjugate of a 2-cycle by a 4-cycle is another 2-cycle. Hence the Sylow 2-subgroup(s) are isomorphic to either C4 × C2 or D4. Suppose the former, then this subgroup would contain four 4-cycles and three 2-cycles by 2-cycles (the square of a 4-cycle is of this second type). But this is impossible because C4 × C2 is Abelian whilst 4-cycles and 2-cycles by 2-cycles do not commute. Therefore, the Sylow 2-subgroup(s) are isomorphic to D4 (see also (b) on page 156), and it is easily seen that

S4 has three Sylow 2-subgroups isomorphic to D4 in one conjugacy class: (1, 2, 3, 4), (1, 3), (1, 2, 4, 3), (1, 4) and (1, 3, 2, 4), (1, 2). These subgroups are not normal (there is more than one, see Theorem 6.10).

Remaining Subgroups

Next we look for possible subgroups of order 12. If one exists, it would be normal by Problem 2.19. By Theorem 3.11,wehave

the set of even permutations in S4 forms a (normal) subgroup of order 12 which is isomorphic to A4. This is the only subgroup of order 12. A normal subgroup must include the neutral element, and be a union of conjugacy classes (Theorem 2.29), and in this case there are no other possibilities. This can be checked using the table opposite. By Lagrange’s Theorem (Theorem 2.27) possible subgroup orders for S4 are 1, 2, 3, 4, 6, 8, 12, and 24. We have treated the cases 2, 3, 8, and 12; and, as with all groups,

S4 has (normal) subgroups e and S4. Hence we need to consider possible non-cyclic subgroups of orders 4 and 6. Clearly, the permutations in S4 which ﬁx the symbol 1 (or 2, or 3, or 4) form a subgroup isomorphic to S3( D3) (Problem 2.20), hence

S4 has four subgroups isomorphic to S3: (1, 2, 3), (1, 2), (1, 2, 4), (1, 2), (1, 3, 4), (1, 3), and (2, 3, 4), (2, 3).

A non-cyclic group of order 6 is isomorphic to S3 (Problem 2.20), and so this is the only possibility. Finally, we note that S4 has nine elements of order 2, and so, potentially, it has a number of subgroups of the type C2 × C2 T2. First, we note that e and the three elements in the second conjugacy class form a normal subgroup which we denote by V , see page 19,so

S4 has a normal subgroup isomorphic to: C2 × C2 : (1, 3)(2, 4), (1, 4)(2, 3) = V. (8.1) 168 8 Groups of Order 24—Three Examples

No other full symmetric group (apart from S4) has a proper non-neutral normal subgroup which is not an alternating group. After some further checking, we have, noting that (i, j)(i, k) = (i,j,k)if i, j and k are distinct,

S4 has three non-normal subgroups isomorphic to C2 × C2 which form a single conjugacy class: (1, 2), (3, 4), (1, 3), (2, 4) and (1, 4), (2, 3).

As we have considered all possibilities our list of subgroups of S4 is complete.

Centre As Z(S4) is a normal Abelian subgroup of S4 (Lemma 2.31), it follows that the centre can only be V or e,see(8.1) and the diagram opposite. But

(1, 2)(3, 4) · (1, 3) = (1, 2, 3, 4) = (1, 4, 3, 2) = (1, 3) · (1, 2)(3, 4), for instance, and so at least one element of V does not belong to the centre. There- fore, Z(S4) =e, and because of this the group S4 is called centreless.

Derived Subgroups The conjugate of a j-cycle is another j-cycle (Theorem 3.6), and so the commutator of two elements in S4 is an even permutation and belongs to A4. Also we have, for distinct i, j, k ∈{1, 2, 3, 4}, (i, j), (i, k) = (i, j)(i, k)(i, j)(i, k) = (i,j,k)2 = (i,k,j); that is, every 3-cycle belongs to the derived subgroup, hence S4 A4. For the second derived subgroup, we have (i,j,k),(i,k,l) = (i, l)(j, k). By Theorem 3.12, this shows that (S4) = V ,see(8.1). The third derived subgroup is e because V is Abelian.

****** The diagram given on the opposite page illustrates the structure of the lattice of subgroups for S4. The left-hand column gives the subgroup type, the second column gives the order, and the right-hand column gives the number of subgroups. The symbol ‘N’ indicates that the corresponding subgroup is normal in S4. The subgroups are circled and the lines between the circles correspond to the subgroup relation working upwards; so, for example, (1, 2, 3) is a (maximal) subgroup of both (1, 2, 3), (1, 2) and A4. Also the rows in the diagram correspond to subgroup conjugacy classes. To simplify the diagram slightly, the linkages with the six subgroups generated by single 2-cycles are indicated by arrows and the letters A,B,...,F. For instance, the far left-hand downward arrow with A,D,E below it indicates that the subgroup generated by (1, 2, 3) and (1, 2) has itself three subgroups of order 2 (they are (1, 2) labelled E, (1, 3) labelled A and (2, 3) labelled D). The far right-hand upward arrow labelled F indicates that (3, 4) is a subgroup of two of the subgroups isomorphic to S3 on the left-hand page ((1, 3, 4), (1, 3) and (2, 3, 4), (3, 4) again with the label F ). Note that (3, 4) is also a subgroup of (1, 2), (3, 4) with the label E,F below it. 8.1 Symmetric Group S4 169 170 8 Groups of Order 24—Three Examples

Central Series The lower Di(G) and upper Zi(G) central series of a group G are deﬁned in Section 10.1.ForS4 the lower series has two terms: D1 = S4 and D2 = Z = S4 A4, and the upper series only has one term 0 e because S4 is centreless. S4 is the only group of order 24 whose lower and upper series have different lengths, and this shows that the group is not nilpotent.

Frattini Subgroup This subgroup is deﬁned as the intersection of all maximal subgroups of the group in question, see Section 10.2. The maximal subgroups of S4 include (2, 3, 4), (2, 3), (1, 2, 3, 4), (1, 3) and (1, 3, 2, 4), (1, 2); as these have no common element except e we deduce that the Frattini subgroup (S4) =e.

Fitting Subgroup and Radicals The group S4 has only four normal subgroups and the Fitting subgroup (Section 10.2) must be one of these. It is also nilpotent, and so it cannot be isomorphic to either A4 or S4, see Theorem 7.9. Hence by maximality, the Fitting subgroup F(S4) = V C2 × C2.Thep-radical Op(S4) is deﬁned as the intersection of the Sylow p-subgroups, see Problem 6.9(vi). In S4,wehave O2(S4) = V and O3(S4) =e.

Automorphisms The basic properties were given in Section 4.4 and by Corol- lary 5.27.AsS4 is centreless the subgroup of inner automorphisms Inn S4 is isomorphic to S4 (Corollary 5.27). The group has no outer automorphisms, see Prob- lem 8.5, and so the full automorphism group of S4, that is, Aut S4, is isomorphic to S4 itself, and all automorphisms are given by conjugation.

Representations It was first shown by von Dyck in 1882 that S4 has the presentation 2 3 4 G1 = a,b | a = b = (ab) = e S4, (8.2) see Problem 3.18. To prove this we set a → (3, 4) and b → (1, 2, 3), then ab → (1, 2, 3, 4), and a,b and ab satisfy the relations in (8.2). Hence G1 is a homomorphic image of S4, and it is a straightforward exercise to show that it contains 24 distinct elements, therefore it is isomorphic to S4. The group has several other easily defined presentations, one using involutions was given in Problem 3.21. Another is | 3 = 3 = 4 = 2 = 2 = 2 = 2 = S4 a1,b1,c1 a1 b1 c1 (a1b1) (b1c1) (c1a1) (a1b1c1) e see Problem 8.5(ii). This presentation is sometimes denoted by G3,3,4 where the general group Gr,s,t is defined as above except that the powers 3, 3 and 4 are replaced by r, s and t, respectively. It arose during the study of the projective general linear group PGLn(q), that is, the general linear group GLn(q) factored by 3,5,5 its centre; see Coxeter and Moser (1984), page 96. We also have A5 G and 3,7,8 PGL2(7) G which has order 336. The group S4 can also be treated as the rotational symmetry group of a (regular) octahedron as follows. An octahedron has eight identical equilateral triangular faces, and so it has three types of symmetry: (i) rotation by π about a line through the 8.1 Symmetric Group S4 171 centre points of opposite edges, (ii) rotation by 2π/3 about a line through the centres of opposite faces and (iii) rotation by π/2 about a line through opposite vertices. To show that this group is isomorphic to S4, use the presentation given in (8.2), and associate a rotation of type (i) with a, a rotation of type (ii) with b, and a rotation of type (iii) with ab. The reader should check that a rotation of type (i), followed by a rotation of type (ii), has the same effect on an octahedron as a rotation of type (iii). For this reason some authors call S4 the octahedral group. Note also that a cube can be inscribed in an octahedron (with its vertices corresponding to the centres of the faces), and vice versa; and so S4 can equally be treated as the rotational symmetry group of a cube. But the full symmetry group (that is, if reflections are also counted) of an octahedron or a cube is S4 × C2. This can also be represented by a wreath product of the form C2 wr S3 (page 156). There are a number of representations of S4 as a matrix group. In Problem 3.12, we showed how a permutation group can be represented as a group of ‘permutation’ matrices, and in Problem 12.6 we prove that S4

2 3 Remember that 1 + 1 = 0inF2. It is easily checked that C = AB, A = B = 4 C = I3,the3× 3 identity matrix; and so A and B generate an isomorphic copy of S4 in GL3(2). Second, if we take the group GL 2(3) and factor out its centre (which 20 has order 2 and is generated by the matrix 02 ), we obtain another group of order 24 usually denoted by PGL2(3) which is isomorphic to S4; see page 170. S4 is not decomposable (it cannot be written as a direct product with non-neutral factors), but it has two representations as semi-direct products as follows:

S4 C2 A4 if we take C2 =(1, 2) and A4 to be the set of even permutations in S4; we could replace (1, 2) by a subgroup generated by another single 2-cycle in S4; reader, why?

S4 S3 T2 if, for example, we take S3 to be the subgroup of permutations on 1, 2 and 3 only, and T2 = V ,see(8.1). Finally, note that, for groups of order 24, each of the following statements characterise S4 (these are not the only characterisations, for instance, see the paragraph on central series above): The group S4 is the only group of order 24 (Appendix C) such that (a) it has no normal subgroups of orders 2 or 3, (b) it is the only group with no normal Sylow subgroups, that is, with both n2 > 1 and n3 > 1, (c) it is centreless—its centre is the neutral subgroup, (d) it has the largest derived subgroup, (e) it occurs as a subgroup of A6, see Problem 3.7 and Cayley’s Theorem (Theo- rem 4.7). 172 8 Groups of Order 24—Three Examples

The group S4 is also the largest soluble symmetric group, and it occurs as a maximal subgroup of several simple groups; see the ATLAS (1985).

8.2 Special Linear Group SL2(3)

For our second example, we consider SL2(3), the group whose initial definition is as the set of all 2 × 2 matrices with determinant 1 defined over F3, the 3-element field. By Theorem 3.15, it has 24 elements. We shall see below that this group can be defined in several ways (that is, like most groups it has several distinct representations), but we introduce the group using direct matrix calculations via a suitable computer algebra package such as GAP. The elements with their orders are tabulated below. A further series of straight-forward calculations shows that the conjugacy classes of SL2(3) are given by the rows in this table. An important point to note is that this group has only one involution. Cauchy’s Theorem (Theorem 6.2), or Problem 2.8, implies that at least one involution must occur as the group has even order, in this case there are no more.

Number Element List of elements of elements order

1110 01 1220 02 4311 , 10 , 21 , 01 01 21 20 22 4310 , 12 , 22 , 02 11 01 10 12 6422 , 12 , 21 , 11 , 02 , 01 21 22 11 12 10 20 4621 , 20 , 01 , 11 02 22 21 20 4620 , 22 , 02 , 12 12 02 11 10

Next we look for a presentation of the group, one is as follows. If we let 21 20 a → and b → , (8.3) 02 12

ab → 22 ba → 12 then 21 , 22 and 20 a3 = b3 = (ab)2 → , 02 the unique element of order 2. 8.2 Special Linear Group SL2(3) 173

Lemma 8.1 If 3 3 2 G2 = a,b | a = b = (ab) , then G2 SL2(3).

Proof We begin by showing that

a6 = b6 = e. (8.4)

By cancellation, the relations in G2 give

a2 = bab and b2 = aba, (8.5)

and using (8.5) several times we obtain

a6 = a2 3 = bab2ab2ab = ba2ba3ba2b = ba2b5a2b = ba2b6ab2 = ba9b2 = b12 = a12,

and so (8.4) follows by cancellation. The maps in (8.3) and the table above show that G2 is isomorphic to a factor group of SL2(3), hence to prove the lemma we need to show that o(G2) = 24. This can be done directly using (8.4) and (8.5), we leave this as an exercise, see below. For instance, (ab)3 = ababab = a4b = b3ab = b2a2.

This result shows that the elements of G2 are e; a3; a2,b4,b2a,ab2; b2,a4,ba2,a2b; ab,ba,a2b2,b2a2,ab2a,ba2b; a,b5,a5b,ba5; b,a5,ab5,b5a; where the orders of the elements in this list correspond to those given in the table opposite starting with e in row 1.

Subgroups of SL2(3)

The group SL2(3) has ﬁfteen subgroups in seven conjugacy classes. As in the previous example, we begin by listing the cyclic subgroups. The group has a single element of order 2, so SL ( ) 20 a3 2 3 has a single cyclic subgroup of order 2: 02 . The group has eight elements of order 3 (use table opposite), and so SL (3) has four cyclic subgroups of order 3: 11 a2, 10 b2, 2 01 11 21 b2a 22 ba2 20 , and 10 , 174 8 Groups of Order 24—Three Examples see the Sylow entry below. Similarly, as there are six elements of order 4, the group contains three cyclic subgroups of order 4; the intersection of any two of which equals the unique 2-element cyclic group: SL (3) has three subgroups isomorphic to C : 22 ab, 2 4 21 21 a2b2 02 ab2a 11 , and 10 . Finally, as there are eight elements of order 6, we have SL (3) has four cyclic subgroups of order 6: 21 a, 20 b, 2 02 12 01 a5b 02 ab5 21 , and 11 ,

20 and again each contain the matrix 02 .

Sylow and Remaining Subgroups

The four Sylow 3-subgroups are listed above. As in the S4 case, the group has four non-normal cyclic Sylow 3-subgroups. Now SL2(3) has a single element of order 2. The group C8 cannot be a Sylow 2-subgroup (because SL2(3) contains no elements of order 8), and of the remaining groups of order 8, Q2 is the only one with a single involution, hence

SL (3) has a single subgroup isomorphic to Q which is generated by 22 2 2 21 12 ab ba and 22 ,thatis,by and ; it is normal (by Sylow 4).

No other subgroups exist apart from those listed above, e and SL3(2) itself. By Lagrange’s Theorem (Theorem 2.27) and the Sylow theory we only need to consider non-cyclic subgroups of orders 4, 6 and 12. But each of these groups, that is, C2 × C2, D3, and the ﬁve groups of order 12, have more than one element of order 2, hence they cannot be isomorphic to a subgroup of SL2(3).

3 3 3 Centre As a (= b ) commutes with both a and b,wehavea ≤Z(SL2(3)).In fact, we have equality because the centre is both normal and Abelian, and SL2(3) has no other normal Abelian subgroup.

Derived Subgroups Using the presentation given in Lemma 8.1, and (8.4) and (8.5), we have

[a,b]=a5b5ab = a2b2ab = a2ba2 = ab2a,

2 and similarly [b,a]=ba b. Hence (SL2(3)) equals either ab,baQ2 or the group itself. But as SL2(3)/Q2 is Abelian (it has order 3), we have (SL2(3)) ≤ Q2 (Problem 4.6(ii)) and so these groups are equal. In Problem 6.3,weshowed 3 that (SL2(3)) =a , and so the third derived subgroup is e. Also note that Z(SL2(3)) = (SL2(3)) . 8.2 Special Linear Group SL2(3) 175

The diagram above illustrates the subgroup structure of SL2(3)—note there is no subgroup of order 12. The ﬁrst left-hand column gives the subgroup type, and the second column gives the corresponding order. As before the symbol N indicates that the subgroup in question is normal in the group, and the rows list the members of the subgroup conjugacy classes.

******

Central Series Both the lower and upper central series of SL2(3) have two terms and the group is not nilpotent, see Section 10.1:

Lower series: D1 SL2(3) = SL2(3) and D2 SL2(3) = SL2(3) Q2,

Upper series: Z0 SL2(3) =e and Z1 SL2(3) = Z SL2(3) C2.

Frattini Subgroup This normal subgroup is the intersection of the maximal subgroups of SL2(3). From the diagram above we can see immediately that the Frattini (SL ( )) 20 =Z(SL ( )) subgroup 2 3 equals 02 ( 2 3 ). 176 8 Groups of Order 24—Three Examples

Fitting Subgroup and Radicals The Fitting subgroup is the largest normal nilpo- SL ( ) 21 , 11 Q tent subgroup of 2 3 . This is clearly 11 12 (isomorphic to 2) because it is a 2-group and so nilpotent, see page 213. Note that in this case the Fitting subgroup equals the Sylow 2-subgroup. The 2-radical O2(SL2(3)) Q2, and so equals the Fitting subgroup, and O3(SL2(3)) =e.

Automorphisms As o(Z(SL2(3)) = 2, the subgroup of inner automorphisms Inn SL2(3), has order 12 (Corollary 5.27), and is isomorphic to A4. The reader should check this by considering the factor group given by SL2(3)/Z(SL2(3)),see Problem 4.4(iv). The group also has an outer automorphism φ which can be de- ﬁned by interchanging the generators a and b; that is aφ = b and bφ = a; note that by Lemma 8.1 the relations of this group are symmetric in a and b.Aftersome further checking (using Theorem 3.6) we see that the full automorphism group is isomorphic to S4 generated by the inner automorphisms and φ.

Representations We introduced SL2(3) as a matrix group, we also gave a presentation in Lemma 8.1. It has a second presentation given by | 3 = = SL2(3) a1,b1 a1 e,a1b1a1 b1a1b1 , see Problem 8.6(i). The group is a subgroup of S8 (and not of S7—this can be checked theoretically, see Problem 8.6(ii), or by using a computer search), and so it can be represented as a permutation group as follows: SL2(3) (1, 2)(3, 5, 7, 4, 6, 8), (1, 3, 6, 2, 4, 5)(7, 8) ≤ S8.

Note that if the elements in the permutation representation above are denoted by a and b we have a3 = b3 = (1, 2)(3, 4)(5, 6)(7, 8) = c, an element of order 2, and 2 2 (ab) = ((1, 4, 2, 3)(5, 8, 6, 7)) = c, see Lemma 8.1. Further, SL2(3) is sometimes called the binary tetrahedral group because if we factor it by its centre (of order 2) we obtain a group isomorphic to A4 which as we have seen can be represented as the symmetry group of the tetrahedron, see Problem 3.10. The group also has a representation as a semi-direct product:

SL2(3) C3 Q2 if we take C3 to be one of the Sylow 3-subgroups, and Q2 to be the unique Sylow 2-subgroup.

Finally, note SL2(3) can be described as the only group of order 24 which (a) has no subgroup of order 12; (b) has the property that its derived and Sylow 2-subgroups are equal; (c) has the smallest sockel (product of minimal normal subgroups), see (17) on page 293. 8.3 Exceptional Group E 177

8.3 Exceptional Group E

For our final example, we consider what we call the exceptional group E of order 24. It is not a member of some easily recognisable class, but it can be defined as a semi- direct product in several different ways and it has an important connection with the alternating group A7; see the representation subsection below. It does also appear once a list of all groups of order 24 is sought (Problems 8.10 and 8.11), and it has a kind of ‘symmetry representation’. We define the group E by the presentation E = a,b,c | a4 = b2 = c3 = e,bab = a3,bcb = c,aca3 = c2 . (8.6)

As a and b generate a copy of D4 and c generates a copy of C3, the group E can be treated as a semi-direct product C3 by D4 as cC3 is normal in E. We shall see later that it has at least four distinct and easily deﬁned representations as semi-direct products. We show ﬁrst that o(E) = 24 using the following

Lemma 8.2 The element a2 ∈ E commutes with both b and c, and so belongs to the centre of E.

Proof The last relation for E in (8.6) can be rewritten as ac = c2a, and so

a2c = aac = acca = c2aca = c4a2 = ca2.

Two applications of the ﬁrst relation in (8.6)giveba2 = a2b, and the result follows from these two equations.

Using this lemma, we can easily derive the following equations which we use when working with the group E.

cb = bc, ba = a3b, ba2 = a2b, ba3 = ab, (8.7) ca = ac2,ca2 = a2c, ca3 = a3c2, c2a = ac, c2a2 = a2c2,c2a3 = a3c.

Theorem 8.3 o(E) = 24.

Proof Using (8.7), we note that every element of E can be put in the form ar bsct . This shows that o(E) ≤ 24, and to prove equality use the fact that E is deﬁned as a semi-direct product and in this product we have ar bsct = e implies that r = s = t = 0.

A list of elements and conjugacy classes of E is tabulated overleaf where, as in the previous examples, the rows give the conjugacy classes, nine in all. 178 8 Groups of Order 24—Three Examples

Number of Element List of elements elements order

11e 12a2 22b,a2b 62ab,a3b,abc,a3bc,abc2,a3bc2 23c,c2 64a,a3,ac,a3c,ac2,a3c2 26a2c,a2c2 26bc,a2bc2 26bc2,a2bc

Subgroups of E

The group E has thirty subgroups in sixteen conjugacy classes. As in the previous examples (S4 and SL2(3)), we begin by listing the cyclic subgroups using the table above. We have E has nine cyclic subgroups of order 2 in three conjugacy classes (of sizes 1, 2 and 6): a2; b, a2b; ab, a3b, abc, a3bc, abc2, and a3bc2. As only two elements of order three occur in E, that is, c and c2,wehave E has a unique cyclic subgroup of order 3: c. This subgroup is the unique Sylow 3-subgroup, and so by Theorem 6.10, it is normal. There are six elements of order 4, and so E has three cyclic subgroups of order 4: a, ac, ac2 which form a single conjugacy class. Note that the intersection of any two of these subgroups is a2. Lastly, there are six elements of order 6, hence E has three cyclic subgroups of order 6: a2c, bc, a2bc; the ﬁrst is normal and the remaining two form a conjugacy class, use a3bca =a2bc, et cetera.

Sylow and Non-cyclic Subgroups

The unique Sylow 3-subgroup was given above. For Sylow 2-subgroup(s), we can see directly from the deﬁnition of E that the subgroup generated by a and b is isomorphic to D4, and so by the Sylow theory all Sylow 2-subgroups (and so all 8.3 Exceptional Group E 179 subgroups of order 8) are isomorphic to D4. There are three in all, and they form a single conjugacy class by Sylow 2; hence 2 E has three Sylow 2-subgroups isomorphic to D4: a,b, ac,b, ac ,b. By Lagrange’s Theorem and the Sylow theory, E could also have non-cyclic subgroups of orders 12, 6 or 4; in fact, this is true in all three cases as we show now.

Order 12 There are five (isomorphism types of) groups of order 12 (Section 7.3). Perhaps surprisingly three occur as subgroups of E; cf. the Sylow theory. The group C12 cannot be a subgroup because E contains no elements of order 12. Also A4 cannot be a subgroup because it contains eight elements of order 3, whilst E has only two. On the positive side, we note first that there are eleven elements in E which can be expressed in terms of the generators a and c only, see table opposite; hence, together with e they form a subgroup. Referring to the dicyclic group discussed on pages 59, 130 (for metacyclic representation) and 159, we see that 2 E has one subgroup isomorphic to Q3: a c,aa,c. Secondly, a2 commutes with both b and c (Lemma 8.2), and so also with bc, o(a2) = 2 and o(bc) = 6. Hence 2 E has a subgroup isomorphic to C6 × C2: bc,a . No other element of order 2 commutes with an element of order 6, and so E has no further subgroups of this type. The final possibility is D6. Here again we look for an element x of order 2 and an element y of order 6 which in this case satisfy xyx = y5. We have, noting that o(a2c) = 6 and o(ab) = 2, 2 E has one subgroup isomorphic to D6: a c,ab.

Order 6 The element c and an element of order 2 could form part of a subgroup isomorphic to D3. After some checking we ﬁnd that 3 E has two subgroups isomorphic to D3: c,ab and c,a b which form a single conjugacy class. The cyclic groups of order 6 were listed above.

Order 4 The cyclic case was treated on page 178.Asa2 belongs to the centre by Lemma 8.2 (in fact, Z(E) =a2, see below), and there are a further eight elements of order 2, we have after some checking 2 E has four subgroups isomorphic to the 4-group T2 C2 × C2: a ,b, a2,ab, a2,abc, and a2,abc2; the ﬁrst is normal (note that bab = a2), and the remaining three form another conjugacy class. Having tried all possibilities, the list of subgroups of E is now complete. 180 8 Groups of Order 24—Three Examples

The diagram above illustrates the subgroup structure of the group E. The left-hand column gives the subgroup types; generators, but not relations, are given in the main diagram. For the sake of clarity and simplicity, only one representative of each subgroup conjugacy class is illustrated, the number on the right-hand side of each entry gives the size of the corresponding conjugacy class. If this number is ‘1’, then there is a single subgroup in the class, and it is normal in E (Theorem 2.29). 8.3 Exceptional Group E 181

Finally, we ask which of these thirty subgroups are normal? In fact, nine have this property. A subgroup of index 2 is normal, hence the three subgroups of order 12 just listed are normal, and E and e are normal by deﬁnition. The remaining normal subgroups are obtained by checking the element table on page 178 and noting that a normal subgroup is always a union of conjugacy classes including the neutral class (Theorem 2.29). Hence the remaining normal subgroups are given as follows:

a2 containing classes 1 and 2 (order 2), c containing classes 1 and 3 (order 3), a2,b containing classes 1, 2 and 3 (order 4), and a2c containing classes 1, 2, 4, and 6 (order 6).

The reader should check that the other normal subgroups are also unions of conjugacy classes, and no other subgroup has this property.

Centre Previously we noted that a2≤Z(E) (Lemma 8.2). In fact, equality occurs as we can show now. The centre Z(E) is a normal Abelian subgroup of E that contains a2. Referring to the calculations above and the diagram opposite, we see that if Z(E) = a2 then it can only be a2,b or a2c. Both of these are impossible because neither b nor c belongs to Z(E) as they do not commute with a2. Hence Z(E) =a2.

Derived Subgroups Using (8.7), we have [a,b]=a3bab = a2 and [a,c]2 = 3 2 2 2 2 (a c ac) = c, hence a ,c ∈ E and so a c≤E . On the other hand, Q3 a,c E and E/Q3 is Abelian, which shows by Problem 4.6 that E ≤ Q3.But this also applies to both C6 × C2 and D6, and so E is contained in the intersection of these subgroups, and this is a2c. Hence we have equality, that is, E =a2c. As this subgroup is Abelian E =e.

Central Series The lower and upper central series both have length 3, they are

2 Lower series: D1 = E,D2 = E a cC6 and D3 =cC3, 2 2 Upper series: Z0 =e, Z1 = Z(E) =a C2, and Z2 =a ,bC2 × C2.

Note that D3 =[D2,E] by definition, and so the first line above is easily checked. 2 Also Z2 is defined as the preimage of Z(E/Z1) = Z(E/a ) under the natural 2 3 homomorphism from E to E/Z1, see page 210.WehaveZ(E/a ) a C2; the reader should now check that the preimage is as stated above. As Dn = e (all n), E is not nilpotent.

Frattini Subgroup This subgroup is the intersection of the six maximal subgroups of E. It cannot contain a because a ∈ a2,bc, it cannot contain b because b ∈ a,c, and it cannot contain c because c ∈ ac,b. But each of the six maximal subgroups does contain a2, and so in this case (E) =a2(= Z(E)). 182 8 Groups of Order 24—Three Examples

Fitting Subgroup and Radicals This subgroup is the largest normal nilpotent subgroup of E. Neither Q3 nor D6 is nilpotent (they cannot be expressed as direct products of their Sylow subgroups), but C6 × C2 is nilpotent—it is Abelian so automatically has this property. Hence the subgroup a2,bc is the (unique) Fitting subgroup 2 F(E). Referring to Problem 6.9(vi), we see that the 2-radical O2(E) =a ,b, the intersection of the three Sylow 2-subgroups, and the 3-radical O3(E) =c as the Sylow 3-subgroup is normal. Note that the product of these radicals equals F(E) =a2,bc; see Theorem 10.24.

Automorphisms As in the previous example (SL2(3)), the order of the centre of the group of E is two, and so o(Inn E) = 12. In this case, Inn E D6 (use Corol- lary 5.27). Also as in the previous example, there are outer automorphisms and o(Aut E) = 24. One outer automorphism φ is deﬁned by

aφ = a, bφ = b and cφ = c2.

Automorphisms map elements of order 3 to elements of order 3, and as E has exactly 2 elements of order 3, the map deﬁned above is an automorphism. After some further checking we see that the full automorphism group Aut E is isomorphic to D6 × C2 generated by the inner automorphisms and φ.

Representations As noted at the beginning of this section, the group E has few ‘natural’ representations. It was deﬁned by a three generator presentation, and it is easily seen that it has the two generator presentation: E a,d | a4 = d6 = (ad)2 = (a3d)2 = e (8.8) by setting bc = d in the original presentation. The group E forms a notable subgroup A7. The alternating group A7 has 105 involutions each of which consists of pairs of disjoint 2-cycles, and they form a single conjugacy class. (Note S7 also has 105 involutions each consisting of pairs of disjoint 2-cycles, so use Problem 5.21(iii).) Hence using Theorem 5.19 we see that the centraliser of an A7 involution has order 24, and for each such involution the corresponding centraliser is isomorphic to E. This can be proved as follows. Consider, for example, the involution g = (1, 2)(3, 4). Clearly, a permutation of the set {5, 6, 7} commutes with g, as do the four cycle (1, 3, 2, 4) (note (1, 3, 2, 4)2 = g) and the 2-cycle × 2-cycle (1, 3)(2, 4). Hence as we know the order of the centraliser, it is straightforward to check that it can be taken in the form (1, 3, 2, 4)(6, 7), (1, 3)(2, 4), (5, 6, 7) ≤ A7, (8.9) and if we set

a → (1, 3, 2, 4)(6, 7), b → (1, 3)(2, 4) and c → (5, 6, 7) it is also easily checked that the relations in our original definition of E are satisfied by the permutations a,b and c listed above. Therefore, (8.9) provides a permutation 8.4 Problems 183 representation for the group E. The vital connection between involution centralisers and simple groups is discussed in Chapter 12. Coxeter and Moser (1984, page 110) give a representation of E as the symmetry group of a polygonal figure, but it cannot be realised in three dimensions. The group is also isomorphic to a subgroup of SL2(C) generated by the matrices 01 ω 0 , , −10 0 −ω2 where ω3 = 1 and ω = 1; see the presentation (8.8). A second matrix representation is given by the following 7 × 7 matrices A and D defined over the 2-element field F2 where the dots stand for zeros, see (8.8), ⎛ ⎞ ⎛ ⎞ ...1 ... . 1 ..... ⎜ ⎟ ⎜ ⎟ ⎜ ..1 ....⎟ ⎜1 ...... ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ . 1 .....⎟ ⎜ ...1 ...⎟ ⎜ ⎟ ⎜ ⎟ A = ⎜1 ...... ⎟ and D = ⎜ ..1 ....⎟ . ⎜ ⎟ ⎜ ⎟ ⎜ ....1 ..⎟ ⎜ ...... 1⎟ ⎝ ...... 1⎠ ⎝ .....1 . ⎠ .....1 . ....1 ..

A representation distinct from the above, and also using 4 × 4 matrices deﬁned over F2 is given by Problem 8.7,italsouses(8.8). The group E was deﬁned by a semi-direct product. In fact, it has at least four representations as a semi-direct product as follows:

E D4 C3 by deﬁnition;

E C2 Q3 if we take b for C2 and a,c for Q3; 2 E C2 D6 if we take b for C2 as above, and a c,ab for D6; 2 E D3 T2 if we take c,ab for D3 and a ,b for T2.

Finally, note that E shares many properties with D12 and to a lesser extent with D4 × C3 and D6 × C2, but it is the only group of order 24 which (a) is indecomposable (it cannot be written as a direct product of smaller but non- neutral subgroups) and has three non-isomorphic subgroups of order 12, (b) contains six elements of order 4 and six elements of order 6.

8.4 Problems

Problem 8.1 For each of the groups discussed in this chapter, ﬁnd (a) their largest factor groups, (b) the centralisers of each of their elements, (c) the normalisers of each of their subgroups. 184 8 Groups of Order 24—Three Examples

Problem 8.2 Prove that a group of order 24 possesses a normal subgroup of order 4 or 8. (Hint. Consider Sylow 2-subgroups and their normalisers, and then apply Theorem 5.8.)

Problem 8.3 The group F3,8 is deﬁned by 3 8 −1 2 F3,8 = a,b | a = b = e,b ab = a .

Establish the following facts/properties concerning this group.

(i) o(F3,8) = 24. (ii) All proper subgroups are cyclic, in this case the group is called Frobenius or metacyclic; see Theorem 6.18 and Web Section 6.5. (iii) All of its subgroups except the Sylow 2-subgroups are normal. (iv) F3,8 C8 C3, describe the implied homomorphism (Section 7.3), and ﬁnd the centre and the derived subgroup. (v) Determine the Frattini and Fitting subgroups.

Problem 8.4 Let G be given by G = c,d | c2 = d2 = (cd)3 .

(i) Show that c8 = e. (ii) List the elements of G. (iii) Using the previous problem, prove that G F3,8.

Problem 8.5 (i) Show that Aut S4 S4. Begin by showing that an automorphism preserves 2-cycles if and only if it is inner. = 2 = 2 = = = 2 (ii) Using the substitutions a1 b ,b1 abab ,c1 ab and a c1a1,b a1 show that the second presentation of the symmetry group S4 given on page 170 is valid.

Problem 8.6 (i) Using the permutation representation for SL2(3) given on page 176 show that the second presentation for this group also given on page 176 is valid. 2 4 (Hint. Begin with the substitution a1 = a , b1 = b .) (ii) Show that SL2(3) is not isomorphic to a subgroup of Sn if n<8. One method is as follows. Consider the Sylow 2-subgroups of the groups involved, and note that a Sylow 2-subgroup of S6 is isomorphic to D4 × C2.

Problem 8.7 We have seen that E can be treated as a subgroup of SL7(2), but it is also isomorphic to a subgroup of SL4(2), why? Note that GL4(2) SL4(2) L4(2).

6 2 2 Problem 8.8 (i) Investigate the dicyclic group Q6 =a,b | a = b = (ab) ,see Appendix C. You should determine its subgroups, indicate which are normal and which are maximal, draw a subgroup lattice diagram, ﬁnd the centre, et cetera, and look for possible representations, this is not easy!! See also Problem 3.22. 8.4 Problems 185

(ii) In Appendix C, we state that Q6 has a ‘maximum Cayley count’, that is (via Cayley’s Theorem (Theorem 4.7)) Q6 having order 24 is isomorphic to a subgroup of S24, but it is not isomorphic to a subgroup of Sn if n<24. Complete the following sketch proof of this fact. For Qn to act transitively on m points, m mustbeadivisor of 4n. Secondly, note that S6 contains no dicyclic subgroups. If m = 8, the point stabiliser would have an order divisible by 3 but the only subgroup of Q6 of order 3 is normal, and thus lies in the kernel of the action. Finally, note that a similar argument applies when m = 12, hence the only possibility is m = 24. (This proof was suggested to B. Fairbairn.)

Problem 8.9 Investigate the group A4 × C2, see Appendix C. As in the previous problem, you are asked to ﬁnd its subgroup structure, and look for representations. Note that it is an example of a group which can be represented as both a direct product and a semi-direct product which is not direct.

Problem 8.10 In this and the next problem, you are asked to show that there are exactly 15 (isomorphism classes of) groups of order 24. By the Sylow theory, each group with this order has one or four Sylow 3-subgroups. In this problem, consider groups G with a unique Sylow 3-subgroup. Using the direct and semi-direct product theory, show that there are 12 types. (Hint. Aut(C3) C2. In Section 6.1, we listed the groups of order 8, so you will need to consider homomorphisms mapping these groups to C2, that is, you will need to consider their subgroups of order 4.)

Problem 8.11 Continuing the work in the previous problem, now suppose G has four Sylow 3-subgroups. Using the Orbit–Stabiliser Theorem (Theorem 5.7)show that the normaliser of each Sylow 3-subgroup of a group of this type has order 6, and the intersection of all of these normalisers is a subgroup H of order at most 2. Now use Theorem 5.15. Show that if o(H) = 1 then G S4, and if o(H) = 2 then G is isomorphic to a semi-direct product C3 K where K C2 × C2 × C2 or Q2 as these last two groups are the only ones of order 8 whose automorphism groups have elements of order 3, a fact that you may take for granted; see Section 4.4.You could also consider what is needed to prove this last statement.

Problem 8.12 (Project—Groups of Order 16) Investigate the class of groups of order 16. You should find nine of ‘standard’ type, that is direct products, dihedral, et cetera, and five of ‘exceptional’ or semi-direct type; see Problems 3.20, 3.23, 6.4, 6.23, and 7.20. In each case, list the elements and their orders, the subgroups and their orders, find the centres and the derived, Frattini and Fitting subgroups, and describe any unusual features. They can be characterised by considering their ‘Abelianisation’. For each group G, this means consider G/G; see Problem 4.6, Moody (1994) pages 80 to 84, and Appendix C. Chapter 9 Series, Jordan–Hölder Theorem and the Extension Problem

Continuing our study of group properties, we take a new approach and consider their normal subgroup structure. This will introduce two major classes of groups— nilpotent and soluble—they play a central role in the theory; nilpotent groups will be discussed in Chapter 10 and soluble groups in Chapter 11. But first, in this chapter we prove the Jordan–Hölder Theorem which provides details about a particular series called the composition series. Roughly speaking, this theorem says that we can ‘build’ all finite (and some infinite) groups from simple groups using extensions, and so it demonstrates the importance of simplicity in our quest to discover as complete picture as possible of all groups. We give a brief introduction to extension theory in the second section of this chapter, it extends the work on semi- direct products discussed in Chapter 7. As we shall see, although this theory has developed over many years it does not yet provide a complete account, and further progress here would greatly benefit the whole of group theory. One major result in this area is the Schur–Zassenhaus Theorem, first proved in full generality in 1937. It states:

If K is a normal subgroup of a ﬁnite group G and (o(K), o(G/K)) = 1, then G is isomorphic to a semi-direct product of K by a group A isomorphic to G/K; that is, G can be treated as a particularly simple kind of extension (a ‘split’ extension) of K by A.

More complex extensions arise when o(K) and o(A) = o(G/K) have common factors which, as noted above, will be discussed in Section 9.2. We shall give a proof of this important result in Web Section 9.4, it introduces some elementary aspects of cohomology theory which are needed in the case when the subgroup A is Abelian. Each of these topics uses the notion of a series, that is, a ﬁnite sequence of subgroups of the group in question. Properties of their factors will provide information about these groups. The nilpotent and soluble groups mentioned above have series whose factors have special properties.

H.E. Rose, A Course on Finite Groups, 187 Universitext, DOI 10.1007/978-1-84882-889-6_9, © Springer-Verlag London Limited 2009 188 9 Series, Jordan–Hölder Theorem and the Extension Problem

9.1 Composition Series and the Jordan–Hölder Theorem

We begin with the basic deﬁnitions of a series. Given a group G and a subgroup J of G,aseries from J to G is a ﬁnite sequence

J = H0 ≤ H1 ≤···≤Hn = G (9.1) of subgroups of G where each Hi is a subgroup of its successor. If as is often the case J =e, we say that (9.1)isaseries for G. The subgroups Hi in this series are called terms. The length of the series is the number of terms excluding G itself, that is, the number of steps in the series, n for (9.1). A series is called proper if no two of the terms are equal, that is, Hi

Deﬁnition 9.1 (i) A series is called subnormal if each term Hi in the series is a normal subgroup of its successor; in (9.1) Hi−1 Hi ,fori = 1,...,n. (ii) A subnormal series is called normal if each term in the series is also normal in G;in(9.1) Hi G,fori = 0,...,n. (iii) A proper subnormal series for G is called a composition series for G if it has no proper subnormal reﬁnement. The factors of a composition series are called composition factors.

Notes (a) Normality is not transitive, and so the property given in (ii) is stronger than that given in (i). (b) A reﬁnement of a normal series need not be normal; see Example (a) below. (c) There is no consistent terminology in the literature, some authors call our subnormal series ‘normal’, and our normal series ‘invariant’; and some older books are different again.

Examples (a) Let G = SL2(3) (Section 8.2). The diagram on page 175 shows that 3 e a C2 ab,baQ2 SL2(3) is a normal series for SL2(3). (Reader, what do you notice about the factors of this series?) It has the reﬁnement 3 e a abC4 ab,ba SL2(3). This new series is subnormal and has no further reﬁnements, and so it is a composition series for SL2(3). It is not a normal series because the subgroup ab is not normal in SL2(3). 9.1 Composition Series and the Jordan–Hölder Theorem 189

(b) The group S5 has the series

e A5 S5.

This is a normal series having no proper refinement because A5 is simple, and so it is a composition series for S5. The factors are isomorphic to A5 and C2( S5/A5), that is, the factors are simple groups. (c) One normal series for the group C105 =a (it is Abelian, so all subgroups are normal) is as follows: 21 e C5 a C105 where the factors are isomorphic to C5 and C21, one of which is not simple. But the series has the proper refinement (which cannot be further refined): 7 e C5 C15 a C105.

This is a composition series with simple factors isomorphic to C5,C3 and C7. Fur- ther examples can be found in the (subgroup lattice) diagrams given in Chapter 8.

We begin our development of these ideas by proving some elementary facts about composition series. We say that “K is a maximal normal subgroup of a group G”if K is a proper normal subgroup of G and there is no proper normal subgroup of G which properly contains K, that is, K G, and if K ≤ J G then J = K or J = G. Note that by Deﬁnition 2.12(iii) a maximal subgroup is always proper. We have

Lemma 9.2 (i) Suppose K G. The factor group G/K is simple if and only if K is a maximal normal subgroup of G. (ii) If K1 and K2 are maximal normal subgroups of G and K1 = K2, then K1 ∩ K2 is a maximal normal subgroup of both K1 and K2.

Proof (i) This follows directly from the Correspondence Theorem (Theo- rem 4.16)—if G/K has no proper non-neutral normal subgroup, then G has no proper normal subgroup lying strictly above K, and vice versa. (ii) We use Theorem 4.15, and write H = K1 and J = K2.Wehave H HJ G by Lemma 4.14, and so as H is maximal, either HJ = H or HJ = G.IfHJ = H then J

G/J = HJ/J H/H ∩ J,

by Theorem 4.15.By(i),G/J is simple, so by (i) again, H ∩ J is a maximal normal subgroup of H . We argue similarly for J .

Theorem 9.3 (i) Every ﬁnite group G has a composition series. (ii) All composition factors are simple. 190 9 Series, Jordan–Hölder Theorem and the Extension Problem

Proof By induction on the order of G.IfG is simple, then e G is a composition series with a single simple factor; and so both (i) and (ii) follow. Otherwise G has at least one proper non-neutral normal subgroup. Let K be a maximal normal subgroup of G; and so, by Lemma 9.2(i), G/K is simple. By the inductive hypothesis, K has a composition series all of whose factors are simple. If we add G to the top end of this series for K we obtain a composition series for G which again has all of its composition factors simple. Both parts of the theorem follow.

Result (i) above is not true in general for inﬁnite groups. For example, the group Z does not have a composition series, for by Theorem 4.19 a non-neutral subgroup of Z has the form kZ,forsomek ∈ Z, and this subgroup always has proper non- neutral subgroups, for example, 2kZ, and all subgroups are normal as Z is Abelian.

Jordan–Hölder Theorem

We come now to the Jordan–Hölder Theorem, one of the most important in the ﬁnite theory. In Example (c) on page 189, we gave a composition series for the group C105, it is not unique. For instance, 35 5 e C3 a C21 a C105, is another composition series for C105. In fact, this group has six composition series, but they all have one property in common:

the set of factors, that is, {C3,C5,C7}, is the same in each case, only the order in which they occur changes. It is a remarkable fact that this property holds for all groups with composition series, and so in particular for all finite groups. This example relies on the prime factorisation 105 = 3 · 5 · 7, that is, it relies on the Unique Factorisation Theorem for Z but this latter result can be treated as a corollary of the Jordan–Hölder Theorem. To each group G with a composition series we associate a unique set of simple groups, its composition factors, which can be treated as the ‘building blocks’ for G. This is the Jordan–Hölder Theorem, and explains the emphasis given to simple groups in the theory as a whole although it should be noted that non-isomorphic groups can have the same set of composition factors (Problem 9.2). We shall discuss the Extension Problem—ways in which groups can be built up from their simple composition factors—in the next section. Historically, the French mathematician Jordan showed first that the orders of the factors are fixed, and later the German mathematician O. Hölder (1859–1937) showed that the factors themselves are fixed. In the example above, the factors C3, C5 and C7 are fixed. The example above suggests that we make the following

Deﬁnition 9.4 Two subnormal series for a group G,

e=H0 H1 ··· Hn = G and e=J0 J1 ··· Jm = G, 9.1 Composition Series and the Jordan–Hölder Theorem 191 are called equivalent if n = m, and the set of factors of the ﬁrst series is the same as the set of factors of the second series ignoring the order but counting multiplicities in each series.

See note on isomorphism classes on page 17. It is easily checked (Problem 9.1) that this defines an equivalence relation on the collection of subnormal series for the group G. The Jordan–Hölder Theorem states that two composition series for a group are equivalent. We shall give two proofs, the first is fairly short but applies only to finite groups although it can be adapted to apply to all groups with composition series. This is not a major problem because many of the applications are in the finite case. Our second proof, valid for both finite and infinite groups, requires two preliminary results (Lemma 9.7 and Theorem 9.8) which are of interest in their own right.

Theorem 9.5 (Jordan–Hölder Theorem, Finite Case) All composition series for a ﬁxed ﬁnite group are equivalent.

Proof We use induction on the order of the group G. By hypothesis, we may assume that the result holds for all groups G1 where o(G1)

e=H0 ··· Hn−1 Hn = G and e=J0 ··· Jm−1 Jm = G

be composition series for G.IfHn−1 Jm−1, then the result holds by the inductive hypothesis because o(Hn−1)

Hn−1 ∩ Jm−1 Hn−1 G and Hn−1 ∩ Jm−1 Jm−1 G,

and both of these are composition series from Hn−1 ∩ Jm−1 to G.Also Hn−1Jm−1 = G; this follows using the same method as in the proof of Lemma 9.2(ii) above. Hence by assumption and the Second Isomorphism Theorem (Theorem 4.15), we have

G/Hn−1 Jm−1/Hn−1 ∩ Jm−1 and G/Jm−1 Hn−1/Hn−1 ∩ Jm−1. (9.2) Now if e=K0 ··· Kr−1 Kr = Hn−1 ∩ Jm−1 is a composition series for Hn−1 ∩ Jm−1 (Theorem 9.3), then we have the following four composition series for G, see diagram overleaf:

S1 :e H1 ··· Hn−1 G,

S2 :e K1 ··· Kr−1 Hn−1 ∩ Jm−1 Hn−1 G,

S3 :e K1 ··· Kr−1 Hn−1 ∩ Jm−1 Jm−1 G,

S4 :e J1 ··· Jm−1 G. 192 9 Series, Jordan–Hölder Theorem and the Extension Problem

The diagram illustrates the four series S1,...,S4 used in the ﬁrst proof of the Jordan–Hölder Theorem.

The series S1 and S2 with their last terms (G) removed are composition series for Hn−1, a group with order smaller than o(G). Hence by the inductive hypothesis they are equivalent, and so the series S1 and S2 themselves are equivalent as the ﬁnal factor is the same in each case. A similar argument shows that S3 and S4 are equivalent. Further, the series S2 and S3 are identical except for their penultimate terms, and they have the same length. Also, by (9.2) they have the same set of factors, the only difference being that the last two factors of S2 are interchanged in S3. Therefore, S2 and S3 are equivalent, which shows ﬁnally that S1 and S4 are equivalent by Problem 9.1.

We give two applications here, but note that the theorem’s main signiﬁcance is the central role it imparts to simple groups via the extension problem. First, let us return to Example (b) on page 189 where we showed that

e A5 S5 is a composition series for S5 with factors C2 and A5. The Jordan–Hölder Theorem implies that this series is unique because C2 is not isomorphic to a normal subgroup of S5.Also,S5 cannot have another normal subgroup (other than A5). For if H S5, then e H S5 would be a subnormal series for S5 which could be reﬁned to a composition series for S5. But the factors of this new series must be C2 and A5 by Theorem 9.5, which is impossible unless H is one of these factors. 9.1 Composition Series and the Jordan–Hölder Theorem 193

For our second application, we characterise completely the Abelian simple groups. In Theorem 2.34, we showed that if the order of a group is prime then it is simple and cyclic. Using the Jordan–Hölder Theorem we have the converse; simpler proofs exist but this one is given to illustrate our main theorem ‘in action’.

Theorem 9.6 A ﬁnite simple Abelian group G = e is cyclic with prime order.

Proof As G is simple and G = e, the series

e G (9.3)

is a subnormal series for G with no reﬁnements, and so is a composition series for G.Ifo(G) = mn, m>1, n>1 and (m, n) = 1, then G has a subgroup H of order m (Problem 4.15) and H G because G is Abelian. Hence e H G is a normal series for G which can be reﬁned to a composition series for G one of whose factors is H , or a non-neutral subgroup of H . But this is impossible for by the Jordan–Hölder Theorem all composition series for G are equivalent to the series (9.3). Also, if o(G) = pr and r>0 then, by Theorem 6.5, G has a (proper non-neutral) normal subgroup. Hence o(G) is prime and so, by Theorem 2.34, G is cyclic of prime order.

We end this section by giving a second proof of our main result, one that does not require the group to be finite. It uses the Schreier Refinement Theorem (Theo- rem 9.8) which is proved using an extension of the Second Isomorphism Theorem known as Zassenhaus’s Lemma (or sometimes the Fourth Isomorphism Theorem). It is also sometimes called the ‘Butterfly Lemma’ due to the shape of the diagram on the following page giving the subgroup structure. Zassenhaus introduced his lemma to provide a clearer and more straight-forward proof of Schreier’s Theorem.

Lemma 9.7 (Zassenhaus’s Lemma) Suppose A,A,B and B are subgroups of a group G where A A and B B. (i) A(A ∩ B) A(A ∩ B). (ii) B(A ∩ B) B(A ∩ B). A(A ∩ B) B(A ∩ B) (iii) . A(A ∩ B) B(A ∩ B)

Proof (i) As B B we have, by Problem 2.14(iv), A ∩ B A ∩ B, and as A A this gives A A ∩ B A(A ∩ B)

by Lemma 4.14(iv). This proves (i), and (ii) follows by interchanging A and A with B and B; see diagram overleaf. 194 9 Series, Jordan–Hölder Theorem and the Extension Problem

The diagram illustrates the subgroup structure used in the proof of Zassenhaus’s result, the double lines indicate normality.

Proof of (iii) We use the Second Isomorphism Theorem. We have A∩B≤A(A∩B) and A(A ∩ B) A(A ∩ B) by (i). Also A(A ∩ B) = A A ∩ B (A ∩ B)

because A ∩ B ≤ A ∩ B, and (A ∩ B)∩ A A ∩ B = A ∩ B A ∩ B

by Problem 2.18(i). Using these equations and applying the Second Isomor- phism Theorem 4.15 (with G = A(A ∩ B), H = A ∩ B and K = A(A ∩ B)), we obtain ∩ ∩ ∩ A (A B) = A (A B )(A B) A(A ∩ B) A(A ∩ B) A ∩ B A ∩ B = , (A ∩ B)∩ A(A ∩ B) (A ∩ B)(A ∩ B) see the left-hand side of the diagram above. Applying the same argument we can, using (ii), prove an exactly similar equation with the pair A,A replaced 9.1 Composition Series and the Jordan–Hölder Theorem 195

by B,B, and vice versa throughout, see the right-hand side of the diagram. The last terms of these equations are identical, and so (iii) follows.

Using this lemma we can now prove

Theorem 9.8 (Schreier Reﬁnement Theorem) Suppose

S1 :e=H0 ··· Hn = G, and

S2 :e=J0 ··· Jm = G are subnormal series for G. They have equivalent reﬁnements.

Proof We construct a new subnormal series S1,2 by ‘inserting a copy’ of S2 between each term in S1, and a second new series S2,1 by ‘inserting a copy’ of S1 between each term in S2; we then show that S1,2 is equivalent to S2,1; note that repetitions are allowed. For 0 ≤ r ≤ n and 0 ≤ s ≤ m, let the subgroups Kr,s be given by Kr,s = Hr Hr+1 ∩ Js ,

where we assume that Hn+1 = Hn and Jm+1 = Jm.Wehave Kr,0 = Hr Hr+1 ∩e = Hr and

Kr,m = Hr (Hr+1 ∩ G) = Hr Hr+1 = Hr+1,

as Hr ≤ Hr+1. Further, for 0

Kr,s = Hr (Hr+1 ∩ Js) Hr (Hr+1 ∩ Js+1) = Kr,s+1, using part (i) of Zassenhaus’s Lemma (Lemma 9.7) with Hr = A , Hr+1 = A, Js = B and Js+1 = B. The subnormal series S1,2 can now be deﬁned by

S1,2 :e=K0,0 K0,1 ··· K0,m K1,0

··· K1,m K2,0 ··· Kn,m = G. Interchanging H and J , we deﬁne Ls,r = Js Js+1 ∩ Hr ,

and the subnormal series S2,1 is given by (using (ii) in Lemma 9.7)

S2,1 :e=L0,0 ··· L0,n L1,0 ··· Lm,n = G.

The series S1,2 and S2,1 have the same length (that is, mn + m + n), and using the third part of Zassenhaus’s Lemma (Lemma 9.7), we obtain

K + H (H + ∩ J + ) J (J + ∩ H + ) L + r,s 1 = r r 1 s 1 s s 1 s 1 = s 1,r . Kr,s Hr (Hr+1 ∩ Js) Js(Js+1 ∩ Hr ) Ls,r

This shows that the sets of factors of S1,2 and S2,1 are identical, and so the result follows. 196 9 Series, Jordan–Hölder Theorem and the Extension Problem

We can now reprove the Jordan–Hölder Theorem, the new proof being valid for all groups with composition series.

Second proof of Theorem 9.5 Suppose S1 and S2 are composition series for G, they are subnormal and so, by Schreier’s Reﬁnement Theorem, they S∗ S∗ S S have equivalent reﬁnements, say 1 and 2 .But 1 and 2 are composi- S∗ S∗ S S tion series and so have maximum length, hence 1 ( 2 )is 1 ( 2, respectively) with several repeated terms whose corresponding factors are isomorphic to e. Therefore, S1 and S2 are equivalent composition series.

9.2 Extension Problem

The Jordan–Hölder Theorem states: For groups with composition series, if we can (a) list all simple groups, and (b) solve the extension problem: Given groups K and A G/K, construct all possible non-isomorphic groups G, then we can describe all groups, which is one of the main aims of the theory. Part (a) has been achieved in the finite case, see Chapter 12; hence we need to consider part (b). An answer to (b) has been given by Hölder and Schreier but it has some considerable disadvantages. Given K and A, there may be many solutions G satisfying K G and A G/K, and the Hölder–Schreier theory provides a characterisation of all possible solutions G. But in general it is not possible to determine whether these solution groups G are isomorphic to one another or not. For example, the theory provides p extension groups G of Cp by Cp, but by Theorem 5.22 we know that there are only two non-isomorphic groups of order p2. Nevertheless, the theory has a number of applications and so we shall introduce the basic ideas and consider one case: Cyclic extensions. In principle, this case is sufficient for all soluble groups, see Chapter 11. (An alternative approach has been suggested by Kurosh 1955, Volume 2, pages 202 to 210, which can be sketched as follows. A finite group G has a unique maximal normal soluble subgroup J (Chapter 11), which is characteristic and has the property: G/J has no Abelian normal subgroups. Groups of this type are called semi-simple, their unique maximal normal subgroups are direct products of non- Abelian simple groups. The factor group G/J can be constructed using permutations and automorphisms of simple groups, and the subgroup J has to be constructed using soluble group methods. One advantage of this procedure is that J is unique.) We begin by repeating the definition of an extension.

Deﬁnition 9.9 A group G is an extension of K by A if (i) K G, and (ii) A G/K. 9.2 Extension Problem 197

For example, if G = A K is a semidirect product of K by A, then K G and, using the Second Isomorphism Theorem (Theorem 4.15), we have

G/K = AK/K A/A ∩ K A, that is, G is an extension of K by A. Note that for an extension in general there is no condition exactly replacing the property K ∩ A =e as in the semi-direct product case; but see Problem 9.9. The Schur–Zassenhaus Theorem, see the introduction to this chapter, treats the case when the orders of K and A are co-prime. It can be shown that the collection of all extensions of K by A can be treated as a group (the Second Cohomology Group), and the neutral element of this new group corresponds to the (collection of) semidirect products. For further details, see one or more of the references quoted at the end of this section and Web Section 9.4.

Examples (a) S4 is an extension of A4 by C2, and SL2(3) is an extension of C2 by A4; for both see Chapter 8. 2 (b) The group C4 =a has a unique normal subgroup J =a , and G/J C2. Hence C4 can be treated as an extension of C2 by C2 which is not a semi-direct product because the subgroup J is the unique proper non-neutral subgroup of C4. (c) The matrix group GL2(5) is an extension of SL2(5) by C4, we shall use this example throughout the section, and so for this chapter we call it the Stan- dard Example, that is, we let G = GL2(5), K = SL2(5) and A C4 where we take ∗ (Z/5Z) ={1, 2, 3, 4} with multiplication modulo 5 to be the representation of C4 in this example.

Factor Pairs

Our work on the extension problem begins with the following collection of deﬁni- tions and lemmas.

Deﬁnition 9.10 Given an extension G of K by A and an isomorphism φ : A → G/K,asection of G through A is a set {sa : a ∈ A,sa ∈ G} with the properties

(i) se = e, (ii) sa is a representative of the left coset saK where aφ = saK.

In the Standard Example, there are several choices for the elements of a section. s = I ,s = 21 s = 31 s = 41 (s ) = n One is: 1 2 2 01 , 3 01 , and 4 01 ; note that det n where n = 1, 2, 3 or 4 in this example, that is, n runs through the elements of A. Much of the theory depends on the next two deﬁnitions. As φ is an isomorphism, if a,a ∈ A, then sasa lies in the same coset as saa , and so they differ by an element 198 9 Series, Jordan–Hölder Theorem and the Extension Problem of K ((a, a)ξ in the deﬁnition below). Note also that to start with we are working with an existing extension investigating its construction and properties.

Deﬁnition 9.11 (Part 1) Given G, an extension of K by A, and a section {sa : a ∈ A} of G through A, the function ξ : A × A → K is deﬁned by − = 1 ∈ a,a ξ saa sasa for a,a A.

We have sasa = saa a,a ξ, (9.4) (a, e)ξ = e = (e, a)ξ. (9.5)

Equation (9.4) was given above, and (9.5) follows directly from (i) in Deﬁni- tion 9.10.

Deﬁnition 9.11 (Part 2) (i) Given the function ξ of the extension G with section {sa : a ∈ A} as in Part 1, for a ∈ A the function ϑa : K → K is deﬁned by

= −1 ∈ kϑa sa ksa for k K, that is, kϑa is the conjugate of k by the section element sa. (ii) The pair ξ,{ϑa : a ∈ A} is called the factor pair of the extension G of K by A with section {sa : a ∈ A}.

Note that as ϑa is deﬁned by conjugation, it is an automorphism of K. In the Standard Example, we have (m, n)ξ = I2,ifm or n equals 1 (as s1 = I2), and (m, n)ξ = 11/n ,ifm = 1 = n. Similarly, ϑ is the identity map on SL (5) 01 1 2 and rs − rs r − t(r+ s − t − u)/n ϑ : → s 1 s = (∗) n tu a tu a nt t + u if n = 1; the reader should check these facts. The following two lemmas are basic. To aid clarity in long expressions we have reintroduced the ‘dot’ notation for the group operation in K. This is not strictly necessary and so can be ignored, but it should help in the understanding of the following proofs.

Lemma 9.12 Using the notation set out above, where ξ,{ϑa : a ∈ A} is a factor pair, and a,b,c ∈ A, we have

(a, bc)ξ (b, c)ξ = (ab, c)ξ ((a, b)ξ)ϑc.

This equation is called the cocycle identity. 9.2 Extension Problem 199

Proof We use associativity. We have, applying (9.4) twice, sa(sbsc) = sa sbc(b, c)ξ = sabc (a,bc)ξ (b, c)ξ ,

−1 and, introducing the neutral term scsc and using (9.4)again, −1 (sasb)sc = sab(a, b)ξ sc = sabsc s (a, b)ξsc c = sabsc (a, b)ξ ϑc = sabc (ab,c)ξ (a, b)ξ ϑc

using Deﬁnition 9.11, part 2 for the penultimate equation. The lemma follows by associativity and cancellation.

Referring to the Standard Example, suppose, for instance, a = 2, b = 3 and c = 4. (Remember that in this example A (Z/5Z)∗ ={1, 2, 3, 4} with multiplication modulo 5, the neutral element is 1, and 1/2 = 3, et cetera .) We have ab = 1, bc = 2, (a, bc)ξ = (2, 2)ξ = 13 , (b, c)ξ = (3, 4)ξ = 14 , (ab,c)ξ = (1, 4)ξ = I , 01 01 2 (a, b)ξ = ( , )ξ = 12 ∗ 2 3 01 , and using the identity ( ) on page 198 12 13/4 12 (a, b)ξ ϑ = ϑ = = . 4 01 4 01 01

In this example, the left-hand side of the cocycle identity equals (a,bc)ξ(b,c)ξ = ( , )ξ( , )ξ = 13 14 = 12 2 2 3 4 01 01 01 , and this equals the right-hand side using the displayed equations above because (ab,c)ξ = (1, 4)ξ = I2. The second basic result is

Lemma 9.13 If a,a ∈ A and k ∈ K, then −1 kϑaϑa = a,a ξ kϑaa a,a ξ.

Note this result says that the composition of ϑa and ϑa is deﬁned by conjugation using the ﬁrst term of the factor pair.

Proof Applying Deﬁnition 9.11 twice and (9.4), we have, for all k ∈ K,

− − = 1 = 1 (kϑa)ϑa sa kϑa sa (sasa ) k sasa −1 = saa a,a ξ k saa a,a ξ −1 − = 1 a,a ξ saa ksaa a,a ξ −1 = a,a ξ kϑaa a,a ξ.

Returning to the Standard Example, the expression in the last line of the equation above (call it T ) is a product of three 2 × 2 matrices. 200 9 Series, Jordan–Hölder Theorem and the Extension Problem

= = = For instance, suppose a 2 and a 4, then aa 3 and, using the calculation (a, a)ξ = ( , )ξ = 14 ((a, a)ξ)−1 = 11 k = below Deﬁnition 9.11, 2 4 01 and 01 .If rs ∗ = tu , then kϑaa is given by the right-hand matrix in ( ) on page 198 with n 3. Multiplying these three matrices together gives r + 2t(3r + s + t + 2u)/3 T = . 3t 3t + u

The reader should check that this equals k(ϑ2ϑ4) = (kϑ2)ϑ4 by applying (∗)on page 198 again, ﬁrst with n = 2, and then on the result with n = 4. You should also repeat the whole calculation with different a and a.

Construction of a Group Extension

We come now to the converse of the work described above, that is, given groups K and A, construct G. In this case, we do not have a section to build upon, so we make the following definition as suggested by the lemmas given above. Then using this definition we shall be able to construct G and the corresponding section. As in the semi-direct product case given in Section 7.3, we form a group using the underlying set A × K, and we introduce a new operation by = (a1,k1)(a2,k2) a1a2,(a1,a2)ξ k1ϑa2 k2 , (9.6) where the components of the factor pair ξ and {ϑa : a ∈ A} are given in Defini- tion 9.14 below. This is similar to the semi-direct operation, see page 153, except that we now have the new term (a1,a2)ξ. Or to put this another way, our extension will be semi-direct if (a1,a2)ξ = e for all a1,a2 ∈ A.

Deﬁnition 9.14 Given groups A and K, the maps ϑa : K → K and ξ : A × A → K satisfy the following three properties.

(i) The map ϑe is the identity map on K, and the map ϑa is an automorphism of K for all a ∈ A. (ii) The map ξ satisﬁes (a) (e, a)ξ = e = (a, e)ξ for a ∈ A, and (b) the cocycle equality, that is, for a1,a2,a3 ∈ A, = (a1,a2a3)ξ (a2,a3)ξ (a1a2,a3)ξ (a1,a2)ξ ϑa3 .

(iii) For all k ∈ K and a1,a2 ∈ A,wehave = −1 kϑa1 ϑa2 (a1,a2)ξ kϑa1a2 (a1,a2)ξ.

Using this deﬁnition we can construct the required extension, it depends on the particular automorphisms ϑa chosen. 9.2 Extension Problem 201

Theorem 9.15 Suppose A and K are groups, and ϑa, for a ∈ A, and ξ satisfy the properties given in Deﬁnition 9.14 above. The set A × K with the operation (9.6) forms a group G which is an extension of a group K isomorphic to K by a group A isomorphic to A, and for which the set {(a, e) : a ∈ A} is a section of G through A.

The following proof is straight-forward but somewhat ‘involved’, the reader should write out a version for him(her)self.

Proof The operation is closed by deﬁnition. Next we consider associativity. First, we have using (9.6)twice (a1,k1)(a2,k2) (a3,k3) = a1a2,(a1,a2)ξ k1ϑa k2 (a3,k3) 2 = a1a2a3,(a1a2,a3)ξ (a1,a2)ξ k1ϑa2 k2 ϑa3 k3 . (9.7)

We shall refer below to the argument of the function ϑa3 in this expression, it is a product of three elements of K.Wealsohave (a1,k1) (a2,k2)(a3,k3) = (a1,k1) a2a3,(a2,a3)ξ k2ϑa k3 3 = a1a2a3,(a1,a2a3)ξ k1ϑa2a3 (a2,a3)ξ k2ϑa3 k3 . (9.8) Rewriting the second argument of the pair given in the last expression in (9.8) −1 by introducing the neutral term (a2,a3)ξ ((a2,a3)ξ) gives the expression −1 (a1,a2a3)ξ (a2,a3)ξ ((a2,a3)ξ) k1ϑa2a3 (a2,a3)ξ k2ϑa3 k3. (9.9) Using the cocycle identity (Definition 9.14(ii)) the subexpression between the first set of large round brackets in (9.9) satisfies = (a1,a2a3)ξ (a2,a3)ξ (a1a2,a3)ξ (a1,a2)ξ ϑa3 . (9.10) Also applying (iii) in Definition 9.14 to the subexpression between the second set of large round brackets in (9.9) we obtain −1 = (a2,a3)ξ k1ϑa2a3 (a2,a3)ξ k1ϑa2 ϑa3 . (9.11)

Hence, applying (9.10) and (9.11), and using the fact that ϑa3 is an isomorphism, Expression (9.9) satisﬁes (a1a2,a3)ξ (a1,a2)ξ ϑa (k1ϑa )ϑa k2ϑa k3 3 2 3 3 = (a1a2,a3)ξ (a1,a2)ξ k1ϑa2 k2 ϑa3 k3 202 9 Series, Jordan–Hölder Theorem and the Extension Problem

as ϑa3 is an isomorphism—note its argument, see the comment below (9.7). This expression equals the second argument in the last expression in (9.7); that is, we have established associativity. The neutral element is (e, e), for by (9.6) and as ϑe is the identity map on K (a, k)(e, e) = a, (a, e)ξ kϑe e = (a, k). Further, the inverse operation is given by −1 = −1 −1 −1 −1 −1 (a, k) a , a ,a ξ k ϑa

as the following calculation shows; note that as ϑa is an isomorphism, so −1 is ϑa .Wehaveusing(9.6) and the equation above −1 = −1 −1 −1 −1 −1 (a, k) (a, k) a , a ,a ξ k ϑa (a, k) = −1 −1 −1 −1 −1 e, a ,a ξ a ,a ξ k ϑa ϑa k − = e, a−1,a ξ a−1,a ξ 1 k−1 k = (e, e).

These equations show that the set A × K with operation given in (9.6)formsa group G. The reader should show directly (that is, without using Theorem 2.5 and similar facts) that (e, e)(a, k) = (a, k) and (a, k)(a, k)−1 = (e, e),see Problem 9.13 (for the second equation apply ϑ). Next we prove that G contains a normal subgroup K which is isomorphic to K.LetK ={(e, k) : k ∈ K}.Using(9.6) we obtain for k,k ∈ K (note that ϑe is the identity map on K) (e, k) e,k = e, (e, e)ξ kϑe k = e,kk ∈ K ,

−1 = −1 −1 −1 = −1 ∈ and (e, k) (e, (((e, e)ξ) k )ϑe ) (e, k ) K . For normality (a, k)−1 e,k (a, k) = −1 −1 −1 −1 −1 a , a ,a ξ k ϑa a, (e, a)ξ k ϑe k = −1 −1 −1 −1 −1 e, a ,a ξ a ,a ξ k ϑa ϑa k ϑa k −1 = e,k k ϑa k ∈ K .

An easy exercise shows that K K. Similarly, we deﬁne A ={(a, e)K : a ∈ A} and as in the K case it is clear that A G/K. Lastly, we deﬁne a map ψ : A → G/K by

aψ = (a, e)K for a ∈ A. 9.2 Extension Problem 203

This map is a homomorphism, for if a,a ∈ A aψaψ = (a, e)K a,e K = (a, e) a,e K = aa, a,a ξ K = aa,e e, a,a ξ K = aa,e K = aaψ.

We leave as an exercise for the reader to show that ψ is an isomorphism and that {(a, e) : a ∈ A} is a section of G through A.

Using this theorem we can now obtain a representation for our standard example GL2(5) as an extension of SL2(5) by C4 using the operation described on page 197; the reader should check this.

Cyclic Extensions

As an application of the theorem proved above we consider in more detail extensions of a group by a cyclic group, we noted at the beginning of this section that in principle all soluble groups (Chapter 11) can be generated in this way.

Deﬁnition 9.16 An extension of a group K by a group A is called cyclic if A is cyclic.

Note that the Standard Example is, in fact, a cyclic extension. These extensions are characterised by

Theorem 9.17 (i) Suppose G is an extension of K by a cyclic group A G/K where A =a and o(a) = m. Choose sa ∈ G so that saK is a generator of G/K, m = ∈ and suppose sa k0 K. There exists an automorphism ψ of K with the properties

(a) k0ψ = k0 and m = −1 ∈ (b) kψ k0 kk0 for k K.

(ii) Conversely, given k0 ∈ K and an automorphism ψ of K satisfying (a) and (b) above, the set {(ar ,k): 0 ≤ r

Proof (i) Deﬁne ψ by = −1 ∈ kψ sa ksa for k K. = −1 m = m = m = −m m = −1 Then k0ψ sa sa sa sa k0, and kψ sa ksa k0 kk0. (ii) For the converse, note ﬁrst that in this case ξ only takes the values e r r and k0, and so ξ only depends on values of r. Hence we can put ϑar = ψ , r and we need to show that ξ,ϑar (= ξ,ψ , 0 ≤ rr+ s ≥ m (as t

Therefore, ξ,ϑas , 0 ≤ s

Finally, note that using the hypotheses of the theorem we have (a, e)m = m−1 = m−1 = m m−1 = (a, e)(a, e) (a, e)(a ,e) (a ,(a,a )ξ eϑam−1 e) (e, k0) (because 1 + (m − 1) ≥ m) as required.

An immediate consequence of this result is

Corollary 9.18 If K is a group, g ∈ Z(K) and n is a positive integer, then a group G exists satisfying

n G =K,c,K G, G/K Cn,c∈ CG(K) and c = g.

Proof In Theorem 9.17,letA = Cn and ψ be the identity automorphism on K.

For example, suppose K = C3 and n = 8, then this corollary shows that the group F3,8 with the presentation given in Problem 8.3 does, in fact, exist. Secondly, suppose K is Abelian, g ∈ K, o(g) = 2 and ψ is the automorphism given by aψ = a−1 for a ∈ K, then the corollary asserts the existence of the group G with the properties: − − G =K,c, [G : K]=2,c2 = a, c 1gc = g 1 for g ∈ K.

This group is called generalised dicyclic, and dicyclic if K is cyclic. If in this latter case o(K) = 4, then G Q2 the quaternion group. Reader, referring to Prob- lem 8.12, what group do you obtain if o(K) = 8? We have only been able to give a very brief account of the topics in this section, in particular due to space considerations we have not been able to discuss the basic cohomology theory needed to develop the work. The elementary ideas and results of cohomology theory follow fairly directly from the work of Chapter 2, and Sections 4.1 and 4.2. The reader requiring more details should consult one or more of the following texts: Scott (1964), Chapter 9; Rotman (1994), Chapter 7; Suzuki (1982), Chapter 2; or for a slightly more advanced approach Robinson (1982), Chap- ter 11, or Benson (1991).

9.3 Problems

Problem 9.1 Show that the relation given in Deﬁnition 9.4 is an equivalence relation.

Problem 9.2 (i) Give composition series for the groups: (a) S4,(b)A4,(c)A4 ×C2, (d) SL2(3), and (e) E (Section 8.3). Which groups have unique series? (ii) Give an example of a pair of non-isomorphic groups whose composition series have the same factors.

Problem 9.3 Show that if G has a composition series and K G, then G has a composition series one of whose terms is K. 206 9 Series, Jordan–Hölder Theorem and the Extension Problem

Problem 9.4 (i) Discuss the proposition: An Abelian group has a composition series if, and only if, it is finite. (ii) Show that the cyclic group Cpn has exactly one composition series. (iii) Show that GLn(F ) has a composition series if and only if the field F is finite. (iv) If G has a composition series and H ≤ G, is it necessary for H to have a composition series? (Hint. Refer to Web Sections 3.6 and 7.5.)

Problem 9.5 Prove the following statements. (i) All composition factors of a ﬁnite Abelian group have prime order. (ii) All composition factors of a ﬁnite p-group have order p.

Problem 9.6 (i) Suppose G has a composition series with exactly three terms (length 2). Show that G has a unique composition series or it is isomorphic to a direct product of two simple groups. (ii) Show that groups exist with arbitrarily long composition series.

Problem 9.7 Consider the following two series for Z for primes p and q:

e pZ Z and e qZ Z, where p = q.

Show that these series have equivalent reﬁnements. Schreier’s Theorem (Theo- rem 9.8) asserts the existence of these reﬁnements, but note that it does not require the group in question to have a composition series.

Problem 9.8 A subgroup H of a group G is called subnormal in G if and only if there exists a subnormal series from H to G, and in this case we write H G. Note that unlike normality the subnormal relation is transitive, that is, if J H and H G, then J G. Prove the following three properties. (i) If H G and J ≤ G, then H ∩ J J . (ii) If H G and J G, then H ∩ J G. (iii) If θ : G → G1 is a surjective homomorphism, then H G if and only if Hθ G1.

Problem 9.9 (i) Show that if G is an extension of K by A and both K and A are ﬁnite, then o(G) = o(K)o(A). (ii) Using (i) show that if G is ﬁnite and has the subnormal series e=H0 ··· = = = − = m−1 Hm G, where o(Hi+1/Hi) ri for i 0,...,m 1, then o(G) i=0 ri .

Problem 9.10 (i) Construct all possible extensions of C3 by C2 × C2. (ii) Using Problem 6.5 show that every non-Abelian group with order p3 is an extension of Cp by Cp × Cp.

Problem 9.11 (i) Show that if G and H are Abelian and (o(G), o(H )) = 1, then up to isomorphism there is only one Abelian extension of G by H . (ii) Construct all extensions of the inﬁnite group Z by C2. 9.3 Problems 207

Problem 9.12 Show that the quaternion group Q2 can be written in three different ways as an extension of C4 by C2, and one way as an extension of C2 by C2 × C2. Do any of these extensions reduce to a semi-direct product?

Problem 9.13 In the proof of Theorem 9.15, using only Deﬁnition 9.14 prove that (a) (e, e)(a, k) = (a, k) and (b) (a, k)(a, k)−1 = (e, e).

Problem 9.14 Use extensions to construct all groups of order pq where p and q are prime and p

Problem 9.15 (Project—Chief Series) Before attempting this project, the reader should read the subsection on minimal normal subgroups given at the end of Sec- tion 11.1.Achief series for a group G is a proper normal series

e H1 ··· Hm = G, with the additional property that if K G and, for some i,wehaveHi ≤ K ≤ Hi+1, then either K = Hi or K = Hi+1. This series is proper, so Hi

The material in this chapter is largely disjoint from that given in Chapter 11. Hence with a few minor exceptions these chapters can be read independently. Nilpotent groups lie between the classes of Abelian and soluble groups, this latter class will be discussed in Chapter 11. Finite nilpotent groups also have a number of similarities with p-groups—for example, both have a plentiful supply of normal subgroups; in fact, some authors describe nilpotent groups as ‘generalised p-groups’. In the finite case (which we introduced first in Chapter 6), these groups have many independent definitions showing that they possess many useful properties in common; see Theorem 10.9. Some of these new definitions are ‘series based’ and apply in both the finite and infinite cases. They also provide a ‘measure’—the nilpotency class—of how far away from Abelian a particular nilpotent group is, and this measure can itself be used to establish new results. An Abelian group G has nilpotent class 1, and we have

[a,b]=e for all a,b ∈ G.

For nilpotent groups H of ‘class n’wehave ... [a0,a1],a2 ...,an = e for all a0,...,an ∈ H, see Definition 10.5 and Problem 10.2. One origin of the word ‘nilpotent’ is given in the footnote on page 213. We begin this chapter by stating the series-based definitions of nilpotency mentioned above, in most cases several series are involved and all need to be considered. This builds on the work given at the beginning of Section 9.1. We shall also establish the equivalence of many of the definitions, some of which at first seem unrelated. For instance, one concerns properties of maximal subgroups whilst another states that elements of coprime order commute. In Section 10.2, we discuss the Frattini and Fitting subgroups of a group, in the finite case they are both normal and nilpotent and have some remarkable properties and applications. The first introduces a connection between maximal subgroups and the so-called non-generators of a group, whilst generalisations of the second have proved useful in the CFSG.

H.E. Rose, A Course on Finite Groups, 209 Universitext, DOI 10.1007/978-1-84882-889-6_10, © Springer-Verlag London Limited 2009 210 10 Nilpotency

10.1 Nilpotent Groups

We begin by introducing a new type of series which can be used to define nilpotency in both the finite and infinite cases. The basic series definitions were given at the beginning of Section 9.1.

Central Series

We use the notation [G, H ] for the group generated by all commutators of the form [g,h] where g ∈ G and h ∈ H ; see Problem 2.16. A normal series for G

e=H0 H1 ··· Hn = G, see Deﬁnition 9.1(ii), is called a central series for G if, for each i in the range 0 ≤ i ≤ n − 1, we have

Hi+1/Hi ≤ Z(G/Hi), or equivalently, see Problem 4.16(vii),

[Hi+1,G]≤Hi.

Also G is nilpotent if it has a central series; see Definition 10.5 below. Clearly all Abelian groups are nilpotent, and so nilpotency can be treated as a generalisation of Abelianness. A group can have several different central series, see the example on page 213, and so we begin by studying these series. Previously we have defined the derived (commutator) subgroup G (Prob- lem 2.16) and the centre Z(G) (Definition 2.32) of a group G. Here we extend these notions to define the higher commutator subgroups Di(G) and the higher centres Zi(G), and using these, the lower and upper central series for G. Note the distinction between the higher commutator subgroups given here and the derived subgroups defined on page 234.

Deﬁnition 10.1 The ith higher commutator subgroup Di(G) of G is deﬁned inductively by D1(G) = G, Di+1(G) = Di(G), G . Note that D2(G) = G , the derived subgroup of G, and Di(G) = G for all i if G = G,thatis,ifG is perfect; see Problem 4.8.

Definition 10.2 The ith higher centre Zi(G) of G is defined inductively as follows. Set Z0(G) =e.IfZi(G) (a normal subgroup of G) has been defined, let ηi be the natural homomorphism (Definition 4.13) mapping G to G/Zi(G), then Zi+1(G) is the preimage of Z(G/Zi(G)) under the homomorphism ηi . 10.1 Nilpotent Groups 211

Note that Z1(G) = Z(G) because η0 is the identity map on G and, for i = 0, 1,..., the term Zi(G) is the unique subgroup of G which satisﬁes Z0(G) =e, Zi+1(G)/Zi(G) = Z G/Zi(G) . (10.1)

Also Zi(G) =e for all i when G is centreless. If there exists a ﬁnite integer k such that Zj (G) < Zk(G) for all j

Deﬁnition 10.3 The lower central series for a group G is the series (written in reverse order) from some subgroup of G to G:

G = D1(G) D2(G) ···, and the upper central series for G is the series from e to some subgroup of G:

e=Z0(G) Z1(G) ···.

As observed above, for the lower central series note the similarity of its definition with that for the derived series given on page 234. Problem 10.1 provides a slightly different but equivalent definition of the upper series. When it is clear which group G is involved, we write Dr for Dr (G) and Zs for Zs(G). Both of these series are normal (Definition 9.1); this follows from Prob- lem 4.16 and the Correspondence Theorem (Theorem 4.16). Also neither of these is necessarily a series for G; in the ‘worst’ case Dn(G) = G and Zn(G) =e for all n ≥ 1, for example, when G = S3. The next result shows that there is a close relationship between the lower and upper central series, and in Problem 10.6 we prove that all other central series lie somewhere ‘between’ these two.

Theorem 10.4 If Zs(G) = Zs = G, for some integer s, then Ds+1 =e and

Dr+1 ≤ Zs−r for r = 0,...,s. (10.2)

Conversely, if Ds+1(G) = Ds+1 =e for some integer s, then Zs = G and (10.2) again holds.

Proof For the first part, we use induction on r to prove (10.2). If Zs = G, then by definition of D1, D1 = G = Zs and (10.2) holds when r = 0. If Dr+1 ≤ Zs−r then, using Definition 10.1,wehave

Dr+2 =[Dr+1,G]≤[Zs−r ,G]≤Zs−(r+1),

where the last inequality follows by Problem 4.16(vii) (with G for G, Zs−r for J and Zs−(r+1) for K). Proposition (10.2) now follows and, if we put r = s, we obtain Ds+1 ≤ Z0 =e. 212 10 Nilpotency

For the converse, suppose Ds+1 =e. As above we apply induction, this time on t = s − r to prove D(s+1)−t ≤ Zt . By assumption, we have Ds+1 = e=Z0. Suppose D(s+1)−t ≤ Zt . We use the enlargement of cosets map ξ given in Problem 4.14 with J = D(s+1)−t , K = Zt and ξ = ξt . This provides the surjective homomorphism

ξt : G/D(s+1)−t → G/Zt , (10.3)

and shows that, as D(s+1)−t Ds−t G, Ds−t /D(s+1)−t ξt = Ds−t Zt /Zt . (10.4)

For if a ∈ G, then a(D(s+1)−t )ξt = aZt by deﬁnition, see (10.3) and Prob- lem 4.14, hence if a ∈ Ds−t , the cosets on the right-hand side of (10.4) have the form bZt where b ∈ Ds−t Zt .AlsoD(s+1)−t =[G, Ds−t ] by Deﬁ- nition 10.1, and so by Problem 4.16(vii) again Ds−t /D(s+1)−t ≤ Z G/D(s+1)−t ,

which in turn, by (10.3) and Problem 4.16(vi) (with θ = ξt , G = G/D(s+1)−t , H = Ds−t /D(s+1)−t and J = G/Zt ), gives Ds−t /D(s+1)−t ξt ≤ Z(G/Zt ). Combining this with (10.4), we obtain Ds−t Zt /Zt = Ds−t /D(s+1)−t ξt ≤ Z(G/Zt ) = Zt+1/Zt , using (10.1) for the last equation. From this we have by the Correspondence Theorem (Theorem 4.16)

Ds−t ≤ Ds−t Zt ≤ Zt+1, and the inductive argument is complete. As in the ﬁrst part, if we put t = s, we obtain D1 ≤ Zs and so Zs = G.

Nilpotent Groups

Previously we have discussed finite nilpotent groups, see Sections 6.3 and 7.3.Here we begin by giving a new definition which is valid in both the finite and infinite cases. Using Theorem 10.4,wehave

Deﬁnition 10.5 A group G is called nilpotent with class r if r is the least positive integer satisfying Dr+1(G) =e, and G is called nilpotent if it is nilpotent with class r for some positive integer r.

By Theorem 10.4, we can replace the equation Dr+1(G) =e in the above deﬁ- nition by Zr (G) = G. This theorem also shows that, if G is nilpotent, then the lower and upper central series for G have the same length, see Problem 10.6. 10.1 Nilpotent Groups 213

Example 1 Abelian groups are nilpotent with class 1. Nilpotent groups with class 2 satisfy D2 = G ≤ Z(G) = Z1 (as D2 ⊆ Z(G),wehaveD3 =e). As an example consider the group 8 2 5 G1 = a,b | a = b = e,bab = a discussed in Problem 6.4. It has order 16, so it is a 2-group and hence nilpotent by Theorem 10.6 (or by Theorem 7.9). We also have, using Problem 6.4, D = D = = 4 D = 1 G1, 2 G1 a and 3 e , as a4 is Abelian and a4 ∈ Z(G), and 2 Z0 =e, Z1 = Z(G1) = a and Z2 = G1.

This last equation follows because G1/Z1 is Abelian, and so equals Z(G1/Z1) which in turn equals Z2/Z1 by deﬁnition. Therefore, o(Z2) = 4 · o(Z1), that is, 16; see Problem 6.4 again. These show that G1 is nilpotent with class 2. This also provides an example of a group with distinct lower and upper central series.

Example 2 Consider the groups ITn(Q) and IZTn,r (Q); see Problem 3.15.IfA ∈ ITn(Q) then A is an n × n upper triangular matrix with 1 at each main diagonal entry, and if B ∈ IZTn,r (Q) then B has the same form as A except that the ﬁrst r superdiagonals consist entirely of zeros.1 Using the quoted problem, we have

e IZTn,n−1(Q) ··· IZTn,1(Q) ITn(Q) is a central series for ITn(Q). Therefore, ITn(Q) is nilpotent with class at most n − 1. In fact, the class is exactly n − 1, see Problem 10.2. This example also shows that there exist ﬁnite nilpotent groups with arbitrarily high nilpotency class.

The following results give the basic nilpotency properties, they are derived using straightforward group theoretical techniques and so only outline proofs will be given; the reader should ﬁll in the details.

Theorem 10.6 (i) All Abelian groups are nilpotent (with class 1). (ii) All ﬁnite p-groups are nilpotent (with various classes). (iii) A subgroup of a nilpotent group is nilpotent. (iv) A factor group of a nilpotent group is nilpotent. (v) If G and H are nilpotent, then so is G × H .

1In ring theory, an element a is called ‘nilpotent’ if it is non-zero and some positive power an of it is zero. In the ring of all n × n matrices deﬁned over a ﬁeld F ,ifB ∈ IZTn,r (F ) then B − In is nilpotent in the ring theory sense provided r

The ﬁniteness condition is necessary in (ii) because there exist inﬁnite non-nilpotent p-groups, see the note at the end of this section and Problem 10.3.

Proof (i) If G is Abelian we have Z1 = G. (ii) By induction. Let H be a p-group, if o(H) = p use(i).ByLemma5.21, Z(H) = e, so by the inductive hypothesis H/Z(H) has the series

H/Z(H) H1/Z(H ) ··· Hs/Z(H ) =e,

and then H H1 ··· Hs = Z(H) e is a lower series for H . (iii) By induction, we have H ≤ G implies Di(H ) ≤ Di(G) for all i. Hence Dr+1(G) =e implies Dr+1(H ) =e. (iv) Let θ : G → G/H be the natural homomorphism. By Problem 4.16(vi), Di(G/H )≤Di (G)θ, and so if Dr+1(G)=e then we also have Dr+1(G/H )= e. (v) Again use induction. D1(G × H) = G × H = D1(G) × D1(H ), and Di+1(G × H)=[Di(G × H),G× H]≤[D1(G) × Di(H ), G × H ]≤ [Di(G), G]×[Di(H ), H] by Problem 7.2(i).

An example of a non-nilpotent group is S3 (as Z(S3) =e). This example also shows that the converse of (iii) and (iv) in Theorem 10.6 fails; that is, if H G, and H and G/H are nilpotent, it does not follow that G is nilpotent. In this example, both H = C3 and G/H = C2 are Abelian and so nilpotent. See Theorem 10.16 and Problem 10.7 for partial converses. We have mentioned previously that there are several equivalent definitions for nilpotency in the finite case, we give some more now. The first concerns normalisers and is as follows.

Theorem 10.7 Let G is a ﬁnite group. The following conditions are equivalent: (i) G is nilpotent; (ii) if H

See also Problem 5.13(iv).

Proof We show that (i) implies (ii) implies (iii) implies (i). Suppose (i) holds, we use the equivalent deﬁnition given in Problem 10.1.Letn be the largest integer such that Zn ≤ H .AsH

Example 3 Let G1 = Q2 × C3.BothQ2 and C3 are normal Sylow subgroups of G1 by Lemma 7.3, hence by Theorem 7.9 this group is nilpotent. It has the presentation 2 2 2 3 G1 = a,b,c | a = b = (ab) ,c = e,ac = ca,bc = cb .

Consider H =a2,c, a subgroup of order 6. By Theorem 10.7,wehave

NG1 (H ) > H . This is not surprising for G1 is Hamiltonian, that is, all of its proper subgroups are normal even though it is not Abelian; see Problem 7.13. Hence in this = case NG1 (H ) G1. This problem also shows that all maximal subgroups of G1 are normal, which conﬁrms (iii) in Theorem 10.7 for this group.

Our next equivalent nilpotency condition concerns commuting elements whose orders have no common factors.

Theorem 10.8 A ﬁnite group G is nilpotent if and only if whenever a,b ∈ G and (o(a), o(b)) = 1, then a and b commute.

Proof There is nothing to prove if G is a p-group. Hence suppose the primes dividing o(G) are p1,...,ps where s>1. If G is nilpotent, then by The- orem 7.9, it is a direct product of its Sylow pi -subgroups Pi ,1≤ i ≤ s, say. Re-ordering these factors if necessary, let the prime factors of o(a) be p1,...,pt , and so by hypothesis the prime factors of o(b) are included in the set {pt+1,...,ps}.NowifR1 = P1 ×···×Pt and R2 = Pt+1 ×···×Ps , then G R1 × R2, a ∈ R1 and b ∈ R2, and so a and b commute by Theorem 7.4. For the converse, we can reverse this argument. If for all a ∈ R1 and b ∈ R2, a and b commute, then as in the proof of Theorem 7.4, R1 and R2 are normal subgroups of G, R1 ∩ R2 =e and R1R2 = G (by Theorem 5.8), that is G R1 × R2; the theorem follows.

Example 4 Let G2, a subgroup of S7, be generated by the permutations (1, 2, 3, 4), (1, 3) and (5, 6, 7). The orders of the elements of G2 are 1, 2, 3, 4, 6 and 12, and an easy calculation shows that all of the elements of order 2 or 4 commute with the elements of order 3, hence by Theorem 10.8 it is nilpotent. In fact, it is isomorphic to D4 ×C3, see Appendix C.NowJ =(1, 3), (5, 6, 7)≤G2 with order 6. Therefore,

Theorem 10.7 gives NG2 (J ) > J , as the reader can check directly by showing that = NG2 (J ) (1, 3), (2, 4), (5, 6, 7) , a subgroup of G2 of order 12, thus showing that J is a non-normal subgroup of G2.

The statement on the next page collects together the ten main properties of ﬁnite nilpotent groups, it is important to note that many of these properties fail in the inﬁnite case, see comments on the next page. 216 10 Nilpotency

Theorem 10.9 For a ﬁnite group G, propositions (i) to (x) are equivalent.

(i) G is nilpotent. (ii) For some positive integer r, Dr (G) =e. (iii) For some positive integer s, Zs(G) = G. (iv) If H

Proof Property (v) is discussed in Problem 10.4 and property (x) will be considered in the next section where the Frattini subgroup is deﬁned. For the remaining equivalences, use Theorems 7.9, 10.4, and 10.6 to 10.8.

Properties (iv) to (x) do not apply in the inﬁnite case, they are all too weak; for a discussion of this point see Robinson (1982), page 126. Also there exist inﬁnite non- nilpotent p-groups, an example is given in Problem 10.3. The list of equivalences given in Theorem 10.9 is not complete, four more of slightly less importance are as follows.

(xi) Every factor of a chief series {Hi} for G is central, that is, if the factor is H1/H2 then H2 G and H1/H2 ≤ Z(G/H2). Chief series were discussed in Problem 9.15, and a proof of this equivalence is given in Rose (1978), page 144. This is an example of an equivalence which is valid in both the ﬁnite and inﬁnite cases. The remaining three properties are:

(xii) The factor group G/J is nilpotent when J ≤ Z(G); see Problem 4.16. (xiii) For all proper normal subgroups K of G, Z(G/K) > e. (xiv) For all non-neutral subgroups K of G, [K,G]

Their equivalence to nilpotency is established in Problem 10.9. Lastly, we ask: Is there a condition on n such that all groups of order n are nilpotent? One is as follows. Suppose the prime factorisation of n is given by n = r1 ··· rk p1 pk , then the condition is:

sj − ≤ ≤ ≤ ≤ pi pj 1 for all i, j, sj satisfying 1 i, j k and 1 sj rj .

For example, all groups of order 891 are nilpotent; see Problem 7.12. For a proof of this result, see Scott (1964), page 217. 10.2 Frattini and Fitting Subgroups 217

10.2 Frattini and Fitting Subgroups

Most groups have a number of subgroups with special properties; the centre is often the most important. Here we introduce two further examples: The Frattini and Fitting subgroups. All ﬁnite groups possess these subgroups, but they may be de- generate in the sense that they may equal either the whole group or the neutral subgroup. For instance, if G is Abelian then Z(G) equals G as does the Fitting subgroup of G. Both the Frattini and the Fitting subgroups have a number of useful properties mainly related to the topics discussed earlier in this chapter, we shall introduce them now. First, we consider the Frattini subgroup.

Deﬁnition 10.10 Given a group G, the intersection of all maximal subgroups of G is called the Frattini subgroup of G, it is denoted by (G).IfG has no maximal subgroups we set (G) = G.

Clearly, (G) ≤ G (Theorem 2.15), in fact, it is both normal and (for finite groups) nilpotent as we shall see below (Lemma 10.13 and Theorem 10.15). This subgroup was first defined by G. Frattini (1852–1925) in 1885.

Examples We have (i) (A5) =e, (ii) (C4 C4) C2 × C2, (iii) (C32) = C16, and (iv) (Q) = Q; see Problems 7.20 and 10.13.

There is a second independent deﬁnition which involves ‘non-generators’, the reader should revisit Problem 2.13(ii).

Deﬁnition 10.11 An element a of a group G is called a non-generator of G if whenever the set X generates G, then the set X\{a} also generates G.

For example, the neutral element is a non-generator of all groups whose orders are larger than 1. The ﬁrst result shows that the Frattini subgroup equals the set of non-generators. As with many of the results in this section its proof relies mainly on simple logical arguments involving maximality.

Theorem 10.12 For all ﬁnite groups G, the set of non-generators of G equals the Frattini subgroup of G.

Proof Suppose a is a non-generator of G and H is a maximal subgroup. If a/∈ H , then H,a >H, and so by maximality H,a=G.Buta is a non- generator, and hence H = G which contradicts the maximality of H (note maximal subgroups are always proper). Therefore, a ∈ H , and as this holds for all maximal subgroups H ,wealsohavea ∈ (G). 218 10 Nilpotency

Conversely, suppose a ∈ (G),soa belongs to all maximal subgroups of G; we show that a is a non-generator. Assume that, for some set X, a/∈ X and {a}∪X generates G.IfX =G, then there exists a maximal subgroup J of G which contains X, possibly X itself, that is,

X≤J

By supposition, a ∈ J , and so a,X≤J .Buta,X=G, hence G ≤ J which is impossible. Hence x=G and a is a non-generator.

As noted above, Frattini subgroups have many novel properties, we shall derive some of the main ones now. Four of these apply to all ﬁnite groups whilst the remaining three provide some more information about p-groups. First, we have

Lemma 10.13 (G) G.

Proof If G has no maximal subgroups, then (G) = G and the result holds trivially. An automorphism maps a maximal subgroup to a maximal subgroup, this is a consequence of the Correspondence Theorem. Hence this holds for all inner automorphisms, and normality follows.

The stronger property: (G) char G also holds, see Problem 4.22. The next lemma is needed in the proof of Theorem 10.15, it was ﬁrst proved by Frattini in 1885; cf. Theorem 10.12.

Lemma 10.14 If G is ﬁnite, H ≤ G and G = H(G), then H = G.

Proof If H = G, then there exists a maximal subgroup J of G which contains H possibly H itself. Now (G) ≤ J by deﬁnition, therefore G = H(G)≤ J , which is impossible. The result follows.

Theorem 10.15 If G is ﬁnite then (G) is nilpotent.

Proof We show that all Sylow subgroups of (G) are normal in (G); there is nothing to prove if (G) =e. Suppose (G) > e and let P be a non- neutral Sylow subgroup of (G). Using the Frattini Argument (Lemma 6.14), we have G = NG(P )(G),as(G) G by Lemma 10.13. Now applying Lemma 10.14 with H = NG(P ), we obtain NG(P ) = G, which gives P (G). The result follows as this holds for all Sylow subgroups P of (G).

The next two results provide further conditions for nilpotency; note that in general if H G, and H and G/H are nilpotent, it does not follow that G is nilpotent; S3 is an example as we noted above. 10.2 Frattini and Fitting Subgroups 219

Theorem 10.16 If G is ﬁnite, then G is nilpotent if and only if G/(G) is nilpotent.

Proof If G is nilpotent, then all of its factor groups are also nilpotent by Theorem 10.6(iv). For the converse, we prove that all Sylow subgroups P of G are normal in G as in the proof above. By Problem 6.10(iii) and Lemma 10.13, P (G)/(G) is a Sylow subgroup of G/(G), and as G/(G) is nilpotent by hypothesis, this shows that

P (G)/(G) G/(G)

because all Sylow subgroups are normal in nilpotent groups. The Correspon- dence Theorem now gives P(G) G.AsP is a Sylow p-subgroup of P(G), applying Frattini argument (Lemma 6.14) we obtain

G = NG(P )P (G) = NG(P )(G),

as P ≤ NG(P ). Hence we have G = NG(P ) by Lemma 10.14, that is, P G. As this holds for all Sylow subgroups P of G, the result follows.

We also have: G is soluble if and only if G/(G) is soluble, see Section 11.2. The second new nilpotency criterion completes the proof of Theorem 10.9.

Theorem 10.17 If G is a ﬁnite group, then G is nilpotent if and only if G ≤ (G).

Proof By Theorem 7.9, G is nilpotent if and only if all of its maximal subgroups are normal. If H is maximal, then by hypothesis, H G and H = G, so by Problem 4.13 we see that G/H is cyclic (Abelian). Hence, G ≤ H by Problem 4.6. This holds for all maximal H and so G ≤ (G). Conversely, if G ≤ (G) and H is maximal, then G ≤ H by deﬁnition of (G), and so G H by Problem 2.14(i). It follows that H/G ≤ G/G, but as G/G is Abelian (Problem 2.16), we also have H/G G/G and so, by the Correspondence Theorem, H G. This holds for all maximal H ,soG is nilpotent by Theorem 7.9 again.

A weaker result involving the centre holds for all ﬁnite groups as the following theorem due to W. Gaschütz shows.

Theorem 10.18 If G is ﬁnite, then G ∩ Z(G) ≤ (G).

Proof Suppose G ∩ Z(G) ≤ (G). Then there exists a maximal subgroup H of G which satisﬁes G ∩ (G) ≤ H . Therefore, G = G ∩ Z(G) H, (10.5)

as H is maximal, and so the set G ∩ Z(G) \H is not empty. (10.6) 220 10 Nilpotency

Now if g ∈ G, then g = ah where a ∈ G ∩ Z(G) and h ∈ H (by (10.5)). As a belongs to Z(G) it commutes with all elements in H , and so

g−1Hg = h−1a−1Hah= h−1Hh= H,

which shows that H G. Hence by Problem 4.13, G/H is cyclic and so Abelian, therefore G ≤ H by Problem 2.16.ButG ∩Z(G) ≤ G ≤ H , which contradicts (10.6). Hence G ∩ Z(G) ≤ H for all maximal H , and so ﬁnally G ∩ Z(G) ≤ (G).

Example 5 We illustrate these last two results using the groups discussed in Chap- ter 8. Firstly for S4, both the centre and the Frattini subgroup have order 1 and the derived subgroup is A4, S4 is not nilpotent, and we have equality in Gaschütz’s result. Secondly for Q2 × C3, the derived and Frattini subgroups are equal, and isomorphic to C2, whilst the centre is isomorphic to C6 and contains the ﬁrst two subgroups, Q2 × C3 is nilpotent, and again we have equality in Gaschütz’s result. Finally for F3,8, see Problem 8.3, the centre and the Frattini subgroup are both isomorphic to C4, whilst the derived subgroup is isomorphic to C3, F3,8 is not nilpotent, and we have a strict inequality in Gaschütz’s result.

For p-groups the Frattini subgroup has some useful properties as we show now. See Problem 4.18 for the deﬁnition of elementary Abelian.

Lemma 10.19 If G is an elementary Abelian group, then (G) =e.

Proof In an elementary Abelian group, every non-neutral element can act as a generator, and so the result follows from Theorem 10.12.

Lemma 10.20 If G is a p-group, then G/(G) is an elementary Abelian p-group.

Proof Suppose K is a maximal subgroup of G, then K G and [G : K]=p (Theorem 6.6), hence G/K is elementary Abelian. We use Problem 4.12 to complete the proof. In this problem, let K1,...,Kn be the maximal subgroups of G (and so G/Ki is elementary Abelian for i = 1,...,n). It follows that n G Ki i=1

is also elementary Abelian (a subgroup of an elementary Abelian group is elementary Abelian). The result follows using the deﬁnition of (G).

The last result, which is one of many due to Burnside, has a number of applications, for example, it provides information about the automorphism groups of p-groups; see, for instance, Scott (1964), pages 161 and 162. 10.2 Frattini and Fitting Subgroups 221

Theorem 10.21 If G is a ﬁnite p-group and o(G/(G)) = pn, then we can ﬁnd a1,...,an ∈ G to generate G.

Proof By Lemma 10.20 and Problem 4.18, G/(G) is isomorphic to a direct product of n copies of the cyclic group Cp. Let the generators of these cyclic groups be a1(G),...,an(G), where ai ∈ G for i = 1,...,n. Then

G a1,...,an(G) =a1,...,an using Lemma 10.14 for the last equality.

Example 6 In Problem 6.4, we gave an example of a 2-group G1 with order 16. It 2 2 has three maximal subgroups: a, ab and a ,b, and so (G1) =a with order 4. Hence Burnside’s result conﬁrms the fact that G1 has a 2-element generating set. But if we take the elementary Abelian 2-group of order 16, its Frattini subgroup has order 1 and so the integer n in Burnside’s result is 4, and this elementary group has a 4-element generating set; see Problem 4.18 for details.

Fitting Subgroup

We come now to the Fitting subgroup, it has fewer elementary properties, but generalisations have proved useful in recent work on CFSG. The following result was proved by H. Fitting (1906–1938) in 1938.

Theorem 10.22 (Fitting’s Theorem) If J and K are nilpotent normal subgroups of a group G, with classes r and s, respectively, then JK is also a nilpotent normal subgroup of G with class at most r + s.

Proof The normality of JK follows immediately from Lemma 4.14.For nilpotency we argue as follows. First, note that if Li G for i = 1, 2, 3, then

[L1L2,L3]=[L1,L3][L2,L3] and [L1,L2L3]=[L1,L2][L1,L3]. These identities follow from Problem 2.17 using normality (the reader should verify them). Hence by Deﬁnition 10.1, we have, using the second identity above with L1 = Di(J K), L2 = J and L3 = K, Di+1(J K) = Di(J K), J K = Di(J K), J Di(J K), K . (10.7)

This suggests the following: For subgroups Hi,i = 1, 2,... of G,let[H1]= H1 and [H1,...,Hi,Hi+1]=[[H1,...,Hi],Hi+1] where [G, H ] is as usual the subgroup generated by the commutators of the elements of G and H . Then we have by induction, and using (10.7),

Di(J K) equals the product of all terms of the form [H1,...,Hi], where

Hj = J or Hj = K for j = 1,...,i. 222 10 Nilpotency

Now set i = r +s +1. In the term L =[H1,...,Hr+s+1], J will occur at least r + 1 times, or K will occur at least s + 1 times. Therefore, L is contained in either Dr+1(J ) or Ds+1(K) (Problems 2.16(v) and 10.2(iv)) both of which equal e by hypothesis. This shows that Dr+s+1(J K) =e, and the result follows.

This result leads us to make the following

Deﬁnition 10.23 For a ﬁnite group G, the product of all nilpotent normal subgroups of G is called the Fitting subgroup of G, and it is denoted by F(G).

There is no concensus amongst authors on the notation for the Fitting subgroup, some use F(G),Fit(G) or Fitt(G). By Theorem 10.22, F(G) is the unique largest nilpotent normal subgroup of G; see Appendix C for some examples. Note that by Lemma 10.13 and Theorem 10.15,

(G) ≤ F(G).

Suppose G is a ﬁnite group, p is prime and P1,...,Pk is a list of its Sylow p-subgroups. Then we deﬁne

k Op(G) = Pi. i=1

The (normal) subgroup Op(G) is called the p-radical of G.IfG has only one Sylow p-subgroup P1, then Op(G) = P1 G by Sylow 3. Also, if G has several Sylow p-subgroups (and so not normal), we still have Op(G) G by Problem 6.9(vi). This leads to (for an example, see page 182)

Theorem 10.24 If p1,...,pr are the (distinct) primes dividing o(G), then

= ··· F(G) Op1 (G) Opr (G).

Proof The Sylow p-subgroups of G form a class of conjugate subgroups in G (Theorem 6.9), and so Op(G) is the core (Problem 2.24) of each member of this class, and hence is normal in G. Therefore, as p-groups are nilpotent, Op(G) ≤ F(G) for each p dividing o(G). Now by Theorem 10.22, this shows that r Op (G) ⊆ F(G). i=1 i Conversely, if Q is a Sylow p-subgroup of F(G), then it is contained in some Sylow p-subgroup P of G (Theorem 6.9). By Problem 6.20(i), we have ⊆ = ⊆ r Q G, hence Q core(P ) Op(G), and so F(G) i=1 Op(G).There- sult follows. 10.3 Problems 223

It is not difficult to show that CG F(G) ≤ F(G), provided G is finite and soluble; see Chapter 11 and Kurzweil and Stellmacher (2004), page 123. The solubility condition can be removed if we replace F(G) by ∗ the Generalised Fitting Subgroup F (G) which is defined as the product of F(G) and the subgroup of G generated by the components of G. A component H of G is a subgroup of G which is perfect and subnormal in G (Problem 10.4), and also satisfies the condition H/Z(H) is simple. For example, A5 is a component of G = A5 wr C2; see page 156. The subgroup A5 is perfect, and in G no copy of A5 is normal but A5 A5 × A5 G. Generalised Fitting subgroups play a important role in CFSG. For example, in a series of papers published in the Pacific Journal of Math- ematics during the 1970s John Thompson showed that if G is a non-soluble group ∗ and the normalisers of all of its non-neutral p-subgroups are soluble, then F (G) is isomorphic to one of the following groups most of which will be considered in Chapter 12: 2n−1 2 L2(q), Sz 2 ,A7,M11,L3(3), U3(3), F4(2) , where q>3 and n>1. This means that in most cases (that is, except those listed above whose properties are well known) a non-soluble group will have a non-soluble normaliser attached to at least one of its p-subgroups. For further details, the reader should consult the references quoted above; the Suzuki groups Sz(22n−1) are discussed in Web Section 14.3, and the last group listed is known as the Tits group, it has order 17971200 and some authors (ATLAS 1985, for example) treat it as ‘nearly’ sporadic.

10.3 Problems

Problem 10.1 Show that the following statement gives an equivalent deﬁnition for the upper central series of a group G, see page 211: Z0(G) =e and

Zn+1(G) = g : g ∈ G and [a,g]∈Zn(G) for all a ∈ G .

Problem 10.2 (i) Find the hypercentres of the three groups discussed in Chapter 8. (ii) Show that the dihedral group Dn is nilpotent if and only if n is a power of 2. (iii) Using Problem 3.15, show that the nilpotency class of the group ITn(Q) is exactly n − 1. (iv) Establish the nilpotency condition given in the introduction to this chapter (page 209).

Problem 10.3 For n = 1, 2,...,letHn be a finite p-group with nilpotency class n. Further, let G be the direct product of the Hi , i = 1, 2,.... Show that G is an infinite p-group which is not nilpotent. The definition of infinite direct products is given in Web Section 7.5. 224 10 Nilpotency

Problem 10.4 (i) Show that if H is a subgroup of a nilpotent group G and HG = G, then H = G. (Hint. Use induction on the nilpotency class number and Problem 2.17.) (ii) A subgroup J of G is called subnormal in G if there exists a sequence of subgroups {Ji} with the property

J = J0 J1 ··· Jm = G.

This is expressed by J G. Note that this relation is transitive: if H J and J G, then H G; see Problem 9.8. Prove that G is nilpotent if and only if all subgroups of G are subnormal in G. This establishes the ﬁfth equivalence in Theorem 10.9.

Problem 10.5 (i) Show that if G is nilpotent with class 2 and g ∈ G, then the function θ : G → G deﬁned by aθ =[g,a],fora ∈ G, is a homomorphism, and use it to show that CG(g) G. (ii) Prove that if G is nilpotent with class r, then G/Z(G) is nilpotent with class r − 1. (iii) What is the nilpotency class of the dihedral group D2n ; see Problem 10.2(ii).

Problem 10.6 Suppose G is nilpotent, and e=H0,...,Hn = G is a central series for G; see page 210. Show that Dr+1(G) ≤ Hn−r ≤ Zn−r (G) for 0 ≤ r ≤ n.

Problem 10.7 Suppose G is a group and J ≤ Z(G). Show that G is nilpotent if and only if G/J is nilpotent. (Hint. Use Theorem 10.9(v).) Hall has shown that if K G, and both K and G/K are nilpotent where K denotes the derived subgroup of K, then G is also nilpotent. See Problem 10.11, or for a different proof Robinson (1982), page 129.

Problem 10.8 Suppose G is nilpotent, o(G) = n, and m | n. Show that G has a subgroup of order m; it is reverse Lagrange. (Compare with Problem 7.7).

Problem 10.9 Prove that the statements (xii), (xiii) and (xiv) given on page 216 are, in fact, equivalent to nilpotency. (Hints. For (xii) use Lemma 5.21, and for (xiv) use induction on o(G) and Problem 10.7.)

Problem 10.10 This problem uses the Hall–Witt Identity, see Problem 2.17(iv). Let Hi ≤ G for i = 1, 2, 3, and let [H1,H2,H3]=[[h1,h2],h3]:hi ∈ Hi , i = 1, 2, 3.

(i) Show that if K G, [H2,H3,H1]≤K and [H3,H1,H2]≤K, then

[H1,H2,H3]≤K.

(ii) Prove that if [H2,H3,H1]=e and [H3,H1,H2]=e, then [H1,H2,H3]=e. This result is known as the Three Subgroup Lemma. 10.3 Problems 225

(iii) Using (i) show that if Hi G for i = 1, 2, 3, then

[H1,H2,H3]≤[H2,H3,H1][H3,H1,H2].

(iv) Prove that if G is perfect (Problem 4.8), then G/Z(G) is centreless by applying the Three Subgroup Lemma to two copies of G and one of Z2(G). (v) Lastly, prove that [Di(G), Zj (G)]≤Zj−i(G) if i ≤ j.

Problem 10.11 (i) Show that if G has nilpotency class j, then D[j/2]+1(G) is Abelian. (Hint. Use Problem 10.10(v).) (ii) Suppose K G. Deﬁne Ki and Ln by

n K0 = K, Ki+1 =[Ki,G], and Ln = [Kn−i,Ki]. i=0

Prove that [G, Ln]≤Ln+1. (iii) Using the notation set up in (ii) suppose K has nilpotency class k and G/K has nilpotency class l, show that the nilpotency class of G is at most 2k−1(2l − 1).

Problem 10.12 Suppose G is a nilpotent group. (i) Show that K ∩ Z(G) = e if K is a non-neutral normal subgroup of the group G. (ii) Deduce H ≤ Z(G),ifH is a minimal normal subgroup of G; for the deﬁnition of minimal normal see page 235. (iii) Prove that J = CG(J ),ifJ is a maximal Abelian normal subgroup of G. See also Problem 5.11.

Problem 10.13 Calculate the Frattini and Fitting subgroups of A5, C32, D12 and Sn.

Problem 10.14 Let K G. Show that K ≤ (G) if and only if G has no proper subgroup H which satisﬁes HK = G.

Problem 10.15 (i) If θ is a homomorphism of a group G to itself (that is an endomorphism) show that (G)θ ≤ (Gθ). (ii) Show by an example that the inequality given in (i) can be strict.

Problem 10.16 Prove that if J and K are normal subgroups of G, K ≤ (G) and J/K is nilpotent, then J is nilpotent. (Hint. One method begins by considering Sylow subgroups of J , and using the Frattini Argument (Lemma 6.14) and Lemma 10.14.)

Problem 10.17 Suppose H1,...,Hm are groups. Using Problem 7.3(ii) show that

(H1 ×···×Hm) ≤ (H1) ×···×(Hm).

For a discussion of the reverse inequality, see Scott (1964), page 163. 226 10 Nilpotency

Problem 10.18 Suppose G is a ﬁnite group and K G. Prove the following statements.

(i) If H ≤ G, it does not follow that (H) ≤ (G).(Hint.TryG = S4.) (ii) If H ≤ G and K ≤ (H), then K ≤ (G). (Hint. Assume the contrary and use Problem 2.18 and Theorem 10.12.) (iii) (K) ≤ (G).

Problem 10.19 (i) Let G be a finite group with an Abelian normal subgroup K which satisfies the property: K ∩ (G) =e. Show how to find a subgroup J of G such that G = JK and J ∩ K =e. (Hint. Begin by considering a subgroup J which is minimal subject to the first of these conditions, then use Problem 10.18(ii).) (ii) Give an example from Chapter 8 to illustrate (i).

Problem 10.20 (i) Suppose K G and K ≤ (G). Using Problems 10.15 and 10.18, show that Inn(K) ≤ (Aut(K)). (ii) Prove that if G is ﬁnite, then Inn((G)) ≤ (Aut((G))). (iii) Now applying this result to the dihedral group D4, prove that o((D4)) = 2 and o(Inn(D4)) = 4, and so deduce D4 cannot be the Frattini subgroup of a ﬁnite group. This shows it is not true that every group is isomorphic to a Frattini subgroup of some other group.

Problem 10.21 Suppose G is a ﬁnite p-group. (i) Prove that (G) = GGp where as usual Gp denotes the subgroup of G generated by its pth powers. (Hint. Use Theorem 6.6 and Problem 4.18.) (ii) Use (i) (but not Lemma 10.20) to show that the factor group G/(G) can be treated as a vector space over the ﬁeld Fp.

Problem 10.22 A ﬁnite p-group G is called extra-special if Z(G) is cyclic, and Z(G) = G = (G). (i) Prove that if G is extra-special, then G/Z(G) is elementary Abelian. (ii) Show that if o(G) = p3, then G is extra-special; see Problem 6.5. Further facts concerning these groups can be found in Aschbacher (1994), Chap- ter 1.

Problem 10.23 (i) If G is a 2-group, show that (G) =a2 | a ∈ G. (ii) Show that the proposition given in (i) can be false if 2 is replaced by a larger prime. (Hint. See Problem 6.5.)

Problem 10.24 (i) For a ﬁnite group G show that F(G/(G)) = F(G)/(G). (Hint. Use Problem 10.16.) (ii) Verify directly the result given in (i) for the three groups discussed in Chap- ter 8. 10.3 Problems 227

Problem 10.25 Suppose G is a non-neutral ﬁnite soluble group, see Chapter 11. (i) Show that if e e. (Hint. One proof uses the derived series of G as deﬁned on page 234.) (ii) Deduce CG(F(G)) = Z(F(G)). (Hint. Use terms in the lower central series for a suitably chosen subgroup of G.) (iii) Show that F(G) > e. (iv) Deduce (G) < F(G). Note (ii) shows that the Fitting subgroup of a soluble group plays a similar role to the centre in a nilpotent group, see Problem 10.11.

Problem 10.26 (Project—Supersoluble Groups) A group is called supersoluble if and only if it has a normal series with cyclic factors; for further details, see Scott (1964), Chapter 7. (i) Give an example of a supersoluble group, and another of a group that is not supersoluble. (ii) Using Theorem 10.6 show that we may suppose all cyclic factors of a normal series of a supersoluble group have prime order. (iii) Prove that subgroups and factor groups of a supersoluble group are themselves supersoluble, but the converse can be false. One example is given using S4. (iv) Show that a direct product of supersoluble groups is supersoluble. (v) A finite nilpotent group is supersoluble, and a supersoluble group is soluble; see Chapter 11. Prove these statements and give examples to show that the reverse inclusions are false. (vi) Prove that if G is supersoluble, then G is nilpotent. (Hint. Use the fact that the automorphism group of a cyclic group is Abelian.) (vii) If K is a cyclic normal subgroup of a finite group G and G/K is supersoluble, prove that G is also supersoluble. (viii) The set of odd-order elements of a finite supersoluble group forms a characteristic subgroup. Investigate what is needed to prove this. See Robinson (1982), page 146. (ix) Show that every subgroup of a finite group G is reverse Lagrange if and only if G is supersoluble; see Rose (1978), page 292. (x) Is it true that all Abelian groups are supersoluble? (Hint. Consider what happens in the uncountable case.) Chapter 11 Solubility

See note at the head of Chapter 10. Évariste Galois, who died following a duel at the age of 20 in 1832 (he was involved in French anti-monarchist politics), was one of the main early group theory pioneers. He introduced normal subgroups, soluble groups and some of the results concerning symmetric groups in order to solve a long-standing problem: Can a general polynomial equation be solved by radicals (that is using square-roots, cube- roots, et cetera). He did rely on the work of some earlier mathematicians. Ruffini was the first to claim that some quintic equations are unsoluble, and Abel proved this in the 1820s without making the connection with the permutation group of the roots. Nevertheless, Galois’s contributions, which also included much of the theory of finite fields (Section 12.2), were of the highest originality and importance, and they have had a profound influence on the development of group theory and algebra in general ever since. A brief account of his life and work can be found in van der Waerden (1985)orStewart(1989). It had been known since the sixteenth century that quadratic, cubic and quartic polynomial equations are soluble by radicals, see the references quoted above and Problem 11.1, so why not quintics, sextics, et cetera? The answer depends on the fact to be proved in this chapter that certain groups related to the polynomials in question, S5 amongst them, are not ‘soluble’, whereas others are soluble including Sn for n ≤ 4. Briefly, these groups can be defined as follows; for a more detailed explanation, see any of the standard texts on Galois theory, for example, Artin (1948) or Stewart (1989). Let f(x) be an irreducible polynomial defined over a field F , and let F be an extension field formed by adding one or more of the roots of the polynomial equation f(x)= 0toF . Now consider the set F of automorphisms θ of F which fix F . (If θ ∈ F, then it is a bijection of F to itself, zθ = z for all z ∈ F , and (x + y)θ = xθ + yθ and (xy)θ = xθyθ for all x,y ∈ F .) It is a simple matter to show that F forms a group under composition of maps, and this group is called the Galois group GalF /F (f ) of f for the extension F over F . When the extension field F includes all of the roots of the polynomial equation f(x)= 0, the so-called splitting field for f , the Galois group is denoted by Gal(f ). For example, working over the rational field Q we have, see Cohen (1993), page 327 (where several more examples are given),

H.E. Rose, A Course on Finite Groups, 229 Universitext, DOI 10.1007/978-1-84882-889-6_11, © Springer-Verlag London Limited 2009 230 11 Solubility

5 (a) if f(x)= x − x + 1, then Gal(f ) S5, 5 (b) if g(x) = x − 5x + 12, then Gal(g) D5, and 2 (c) if h(x) = x + x + 1, then Gal(h) C2. In 1829, Abel showed that a quintic polynomial with a commutative (Abelian) ‘Ga- lois group’ is soluble using radicals (that is, using terms some positive power of which belong to the base field F ; see van der Waerden 1985). Galois greatly generalised Abel’s result by showing that in a sequence of extensions with Galois groups G1,G2,..., if each factor group Gi+1/Gi is Abelian, then the corresponding polynomial is soluble by radicals—in Example (c) above, the roots of the polynomial equation x2 + x + 1 = 0 are defined using square roots and the corresponding Ga- lois group is cyclic (Abelian). Repeated applications of these ideas leads naturally to the definition of solubility, see Definition 11.1 below; in fact, the use of the word ‘soluble’ in group theory arises from this result. We begin this chapter by giving the basic facts about solubility. The second section discusses some of the many equivalent definitions including an important one due to Philip Hall, this generalises the Sylow theory but it only applies to soluble groups. In Chapter 10, we studied nilpotent groups, they form an important subclass of the class of soluble groups. For finite groups, we have the (strict) inclusions

Abelian ⊂ Nilpotent ⊂ Soluble ⊂ Generic group.

The class of ‘supersoluble groups’ lies strictly between nilpotent and soluble, see Problem 10.26. Some of these inclusions do not hold for inﬁnite groups. The modern development of solubility theory is covered extensively in Doerk and Hawkes (1992).

11.1 Soluble Groups

We begin with the basic deﬁnition, the reader should refer to the ﬁrst part of Chap- ter 9 for the elementary facts about series.

Deﬁnition 11.1 A group is called soluble1 if it has a subnormal series all of whose factors are Abelian, when this holds the series is called a soluble series.

In Galois’s work on the solution of polynomial equations over fields, the several extensions needed to form the splitting field are mirrored by the steps in the corresponding soluble series. Clearly, all finite Abelian groups are soluble and all non-Abelian simple groups are not soluble. The group S5 is an example which is neither simple nor soluble, see Example (b) on page 189. A wide range of finite groups are soluble including

1Some, mainly American, authors use the word ‘solvable’ for soluble. 11.1 Soluble Groups 231

• all Abelian, nilpotent and p-groups (Theorem 11.6); • all groups whose orders have at most three prime factors (Theorem 11.7), and if we exclude A5 then ‘three’ can be replaced by ‘four’. This count can be further increased to ‘ﬁve’ if a handful of well-known cases are excluded; • all groups which have square-free order, these groups are metacyclic, see Theo- rem 6.18, Problem 6.20 and Web Section 6.5; • all groups of order pr qs where p and q are prime—this is Burnside’s pr qs - theorem which we shall prove in Web Section 14.2; • all groups of odd order—this famous result, which appeared in 1963 and was ﬁrst conjectured by Burnside in the early 1900s, is due to Walter Feit and John Thompson. It is remarkable for its fundamental importance to the theory especially to CFSG, and for its very long proof (about 270 pages)!

We begin our development by showing that subgroups and factor groups of soluble groups are soluble, and vice versa. This is an indication of the importance of solubility in the theory, see also page 116. Note that the constructions and proofs presented in this section rely entirely on material given in Chapters 2 and 4.

Theorem 11.2 A subgroup J of a soluble group G is soluble.

Proof Suppose {H0,...,Hn} is a soluble series for G. We prove the result by showing that

{H0 ∩ J,...,Hn ∩ J } (11.1) is a soluble series for J . It may have some repetitions but this does not affect the result. Note ﬁrst that

H0 ∩ J =e ,Hn ∩ J = J, and Hi ∩ J Hi+1 ∩ J

(Problem 2.14(iv)). Also, for each i we have (by deﬁnition) Hi Hi+1 and Hi+1 ∩ J ≤ Hi+1, and using the Second Isomorphism and Correspondence Theorems we obtain

Hi+1 ∩ J Hi+1 ∩ J as Hi ≤ Hi+1 Hi ∩ J Hi ∩ (Hi+1 ∩ J)

H (H + ∩ J) i i 1 by Theorem 4.15 Hi

H + ≤ i 1 by Theorem 4.16. Hi

This shows that the factors of the series (11.1)forJ are subgroups of the factors of the original series for G, and are therefore Abelian. 232 11 Solubility

Theorem 11.3 If G is soluble and K G, then G/K is also soluble.

Proof Let {H0,...,Hn} be a soluble series for G. By Lemma 4.14, Prob- lem 4.6, and as K G, the series from K to G

K = KH0 KH1 ··· KHn = G

is subnormal. Arguing as in the proof above we have, for i = 0,...,n− 1,

KHi+1 (KHi)Hi+1 = as Hi ≤ Hi+1 KHi KHi H + i 1 by Theorem 4.15 (KHi) ∩ Hi+1 Hi+1 (KHi) ∩ Hi+1 by Theorem 4.17 as Hi,Hi+1 ∩ J ≤ Hi+1. Hi Hi

This shows that KHi+1/KHi is isomorphic to a factor group of the Abelian group Hi+1/Hi , and so is itself Abelian (Problem 4.6). Hence, as KH0 = K, KH KH KH G e = 0 1 ··· n = KH0 KH0 KH0 K is a subnormal series for G/K with Abelian factors (by Theorem 4.17), that is, G/K is soluble.

The converse of these two results is

Theorem 11.4 If K G, and both K and G/K are soluble, then G is also soluble.

This implies that the extension of one soluble group by another is itself soluble.

Proof Let {H0,...,Hn} be a soluble series for K, and let {J0,...,Jm} be a soluble series for G/K. Further, let θ be the natural homomorphism from G to G/K (Deﬁnition 4.13). By the Correspondence Theorem, −1 −1 −1 J0θ ,J1θ ,...,Jmθ

−1 −1 is a subnormal series from K = J0θ to G = Jmθ with Abelian factors. Hence −1 −1 H0,...,Hn−1,J0θ ,...,Jmθ −1 is a soluble series for G because Hn = K = J0θ , the result follows.

In the ﬁnite case, the next result provides a second deﬁnition of solubility.

Theorem 11.5 If G is a ﬁnite group, then G is soluble if and only if it has a composition series all of whose factors are cyclic of prime order. 11.1 Soluble Groups 233

Proof Clearly, if G has a composition series with cyclic factors, then it is soluble as all cyclic groups are Abelian. Conversely, suppose G ﬁnite and soluble. If G is Abelian, the result follows from Lemmas 7.5 and 7.13. For the non-Abelian case, we use induction on the order of G. If the order of G is 1 or a prime, there is nothing to prove. Hence we may assume that (a) o(G) is composite, and (b) the result holds for groups whose orders are less than o(G). Using our supposition, the series

e G

cannot be the only subnormal series for G as this series has a non-Abelian factor. Hence there exists a subgroup K satisfying e K G with 1

We shall now consider some examples. Clearly, all Abelian groups are soluble, as are the dihedral and dicyclic groups; see Problem 11.2. The next two results provide a further wide range of examples.

Theorem 11.6 All ﬁnite nilpotent groups G are soluble.

By Theorem 10.6, this shows that all ﬁnite p-groups are also soluble.

Proof If G is a p-group, this follows directly from Theorem 6.5. Otherwise, by Theorem 7.9, a ﬁnite nilpotent group can be expressed as a direct product of p-groups for various primes p, and the result follows by Problem 11.2.

Theorem 11.7 If G is a ﬁnite group whose order has at most three (not necessarily distinct) prime factors, then G is soluble.

Proof Let p, q and r be distinct primes. There are a number of cases. If o(G) = p,p2 or p3, the result follows from Theorem 11.6.Ifo(G) = pq and p

Using Problem 6.15, this shows that the smallest non-soluble group is isomorphic to A5 with order 60; see Problem 11.3. We noted above that no non-Abelian simple groups is soluble. But groups can be both non-simple and non-soluble as the following result shows. 234 11 Solubility

Theorem 11.8 If n>4, then Sn is not soluble.

Proof The series e An Sn is a composition series for G one of whose factors is not Abelian (Theorem 3.14). Hence by the Jordan–Hölder Theorem (Theorem 9.5), Sn cannot have a subnormal series with Abelian factors, and so Sn for n>4 is not soluble.

Derived Series

The derived subgroup G of a group G is generated by the set of commutators [a,b]=a−1b−1ab where a,b ∈ G.Wehave,ifH G,

G G and G/H is Abelian if and only if G ≤ H, see Problems 2.16 and 4.6. (It is not sufficient just to take the set of commutators in the definition of the derived subgroup. Scott (1964), page 455, gives an example of a group containing two commutators whose product is not a commutator.) Therefore, the series G G can be used as the final step of a soluble series for G, and an extension of this procedure provides an efficient method for constructing this series. We begin with

Deﬁnition 11.9 (i) The nth derived subgroup G(n) of G is deﬁned inductively by: G(0) = G, G(n+1) = G(n) .

(ii) The derived series for G is the normal series from some subgroup of G to G given by ···G(k) G(k−1) ··· G.

The derived length of a soluble group G is the least n such that G(n) =e , see below. Groups with arbitrarily long derived lengths have been constructed; see page 156. Also compare this with Deﬁnition 10.3 for lower central series.

Theorem 11.10 (i) G(n) G. (ii) G is soluble if and only if G(n) =e for some positive integer n.

The proof of this result uses ‘characteristic subgroups’ which were discussed in Problem 4.22. We use this notion because, unlike normality, the characteristic property is transitive, see the quoted problem. Note also G char G.

Proof (i) We use induction on n. By Problem 2.16, G G. Problem 4.22 and the deﬁnition of G(k) give G(k+1) char G(k). Combining this with the inductive hypothesis (that is, G(k) G), we obtain G(k+1) G. Now use Prob- lem 4.22 again. (We also have G(n) char G.) 11.1 Soluble Groups 235

(ii) Suppose ﬁrst G has the soluble series e =Hm ··· H0 = G where for this proof we write the sufﬁxes in reverse order. First, we show

(i) G ≤ Hi for i = 0,...,m, (11.2)

(0) by induction on i.WehaveH0 = G = G , and by Deﬁnition 11.9 and the inductive hypothesis (i+1) (i) G = G ≤ (Hi) .

As Hi/Hi+1 is Abelian by deﬁnition, we also have (Hi) ≤ Hi+1 by Prob- (i+1) lem 4.6. Combining these we obtain G ≤ Hi+1, and (11.2) follows. Now (m) as {Hm,...,H0} is a soluble series for G, this shows that G ≤ Hm =e . Conversely, if G(n) =e for some n ≤ m, then by Problem 4.6 again

e =G(n) G(n−1) ··· G(0) = G

is a soluble series for G.

Minimal Normal Subgroup

Finally, in this section we consider the smallest (minimal) normal subgroups of a soluble group, these subgroups will be used in the proof of Hall’s theorems given in the next section. A subgroup K of a group G is called minimal normal if it is a non- neutral normal subgroup of G, and if J G and e ≤J ≤ K, then either J =e or J = K.IfG is simple, then G itself is the only minimal normal subgroup of G. For example, referring to Section 8.1, the subgroup V generated by a product of two 2-cycles in S4 is an example of a minimal normal subgroup (of S4). In this case, the group S4 is soluble, and the subgroup V (isomorphic to C2 × C2) is elementary Abelian. We show in the next theorem that this is always true for soluble groups. It does not hold in general for non-soluble groups, for instance, consider S5.On page 189, we showed that the only non-neutral proper normal subgroup of S5 is A5, so A5 is the only minimal normal subgroup of S5 and it is clearly not elementary Abelian! See also (17) on page 293 for the sockel, the product of the minimal normal subgrups of the group in question.

Theorem 11.11 If G is a ﬁnite soluble group and H is a minimal normal subgroup of G, then H is an elementary Abelian p-group for some prime p.

Proof Let K be a minimal normal subgroup of G. By Problem 4.22, K char K (where K denotes the derived subgroup of K), and so K G.AsK is minimal, this implies that either K = K or K =e . The ﬁrst possibility is ruled out by Theorem 11.10 as G is soluble; hence K =e , that is, K is Abelian. By Theorems 7.9 and 10.6, K is a direct product of its Sylow subgroups, and each of these subgroups is normal in K. Hence by minimality, K can only have one such subgroup, and so K is an Abelian p-group for some prime p. 236 11 Solubility

p Further, note that if K1 ={k ∈ K : k = e}, then K1 is a characteristic subgroup of the Abelian p-group K (Problem 7.12); and so by minimality again we have K1 = K; that is, K is an elementary Abelian p-group.

It can be shown that if H is a minimal normal subgroup of general ﬁnite group G, then H is either simple or a direct product of isomorphic simple groups; see, for example, Rotman (1994), page 106. This result follows from the fact that if G has no characteristic subgroups (except e and itself), then it is simple or a direct product of isomorphic simple groups. Therefore, Theorem 11.11 shows that in the soluble case these simple groups (factors of the direct product) are always cyclic with prime order.

11.2 Hall’s Theorems and Solubility Conditions

There are a number of equivalent deﬁnitions for solubility (for nilpotency, see Theo- rem 10.9). Perhaps the most remarkable relates to the extension of the Sylow theory to subgroups with coprime orders. In 1928, Philip Hall (1904–1982) showed that if G is soluble and o(G) = mn where (m, n) = 1, then G contains subgroups of orders m and n; and in 1937 he proved the converse, so providing a new characterisation of solubility, one apparently far removed from Galois’s original deﬁnition. We shall prove these results in this section, discuss some related results and applications, and give some further characterisations. We begin with

Theorem 11.12 (Hall’s First Theorem) If G is a soluble group with order mn where (m, n) = 1, then G contains at least one subgroup of order m.

We give two proofs of this result. The first uses the Schur–Zassenhaus Theorem, see page 187, in fact, the proof essentially points out that Hall’s Theorem can be treated as a corollary of the result of Schur and Zassenhaus which appeared in its final form in 1937 nine years after Hall’s first paper. The second proof is a version (based on one given in Rose 1978) of Hall’s original proof; it is quite long and needs careful study, but it does not use any methods not already discussed.

First Proof We use induction on o(G), hence we may assume that the result holds for all groups G1 where o(G1)

Case 2: p m. As above the inductive hypothesis gives H/K ≤ G/K for some H (a subgroup of G) with o(H/K) = m. This implies that o(H) = ms and K H . Now (o(K), [H : K]) = 1, and so the Schur–Zassenhaus Theorem states that H is isomorphic to a semi-direct product of the form H J K where J is a complement of K in H . This result also implies that o(J) = m which completes the proof.

Before giving the second proof we make the following, see also Deﬁnition B.5(ii) on page 285.

Deﬁnition 11.13 Given a set of primes π, a subgroup H of G is called a Hall π-subgroup of G if all the prime factors of o(H) belong to π, and no prime factor of [G : H ] is contained in π.

If π ={p} for some prime p, then a Hall π-subgroup is simply a Sylow p-subgroup. Using Deﬁnition 11.13, we can restate the Hall’s Theorem as:

For all sets of primes π, G contains a Hall π-subgroup.

The main case is when π equals the set of prime factors of m.

Second Proof As in the ﬁrst proof, we use induction on o(G), there is nothing to prove if o(G) is 1 or a prime. Let

Q = Oπ (G).

For the elementary properties of the radical Oπ (G), see Problem 6.9(vi) and Theorem 10.24; the fact it is a normal subgroup can be proved using Lemma 4.14, and Problems 4.11 and 6.9. There are two cases. Case 1: Q>e . As Q G, the factor group G/Q has a subgroup H/Q for some H ≤ G where o(H) is a π-number (by the induction hypothesis). Now o(H) = o(H/Q)o(Q) and so o(H/Q) is a π-number (Deﬁnition B.5). Hence as [G : H ]=[G/Q : H/Q], this shows that [G : H] is a π -number (it only involves primes outside π), so H is a Hall π-subgroup as required. Case 2: Q =e .

Let K = Oπ (G).IfK = G, then G is a π group, there is only one π-subgroup—that is, e , and the theorem follows in this case. Hence we may suppose K

The diagram illustrates the subgroup structure used in the last part of the second proof of Hall’s First Theorem.

****** We apply Problem 9.15. (Note that using Theorem 11.11 most properties of chief series given in this problem are derived in an exactly similar manner to those for composition series given in Section 9.1.) Using this problem we see that G has a chief factor H/K which is an elementary Abelian p-group for some prime p.Wehave p ∈ π. Suppose not, that is, suppose p ∈ π .Aso(H) = o(H/K)o(K), this would imply that H is a normal (the terms in a chief series are normal in G) π -subgroup of G with K

H = PK, (11.3)

for both P and K are subgroups of H , P is Sylow, and K is normal. Fur- ther, as P = e and Q =e ,wehaveNG(P ) < G. (If we had equality, then P G, but this would contradict our assumption.) The inductive hypothesis implies that NG(P ) has a Hall π-subgroup J , see diagram above. We complete the proof by showing that J is a Hall π-subgroup of G, that is, we need to show that [G : J ] is a π -number. By Problem 2.15,wehave [G : J ]= G : NG(P ) NG(P ) : J . (11.4) Also by the Frattini Argument (Lemma 6.14), (11.3) and the fact that a normaliser of a group always contains that group, we have 11.2 Hall’s Theorems and Solubility Conditions 239

G = NG(P )H = NG(P )P K = NG(P )K. Using this and the Second Isomorphism Theorem (Theorem 4.15), we have G : NG(P ) = NG(P )K : NG(P ) = K : NG(P ) ∩ K ,

which is a π -number. Also [NG(P ) : J ] is a π -number by the deﬁnition of J . Combining these facts with (11.4) shows that [G : J ] is a π -number, and the theorem follows.

As in the Sylow theory, it can be shown that two Hall π-subgroups of a ﬁnite group G are conjugate in G, we shall not prove this fact (it is also due to Philip Hall); a proof can be found in Rose (1978), page 285. Hall’s First Theorem cannot be extended to all ﬁnite groups, for example, consider A5 which is not soluble. We have o(A5) = 4·3·5, and A5 does have a subgroup of order 12 (isomorphic to A4) and (12, 5) = 1. But by the example on page 101, A5 does not have subgroups of order 15 or 20 even though (15, 4) = (20, 3) = 1. r1 ··· rn If the prime factorisation of o(G) is p1 pn and G is soluble, then both Hall’s ri and Sylow’s First Theorems state that G has a subgroup of order pi for each i.In the soluble case, Hall’s Theorem extends this to assert the existence of Hall sub- r r groups with order p i1 ···p ik where I ={i ,...,i }⊆{1,...,n}. Perhaps the most i1 ik 1 k important case is when I ={1,...,j− 1,j+ 1,...,n} for some j between 1 and n; in this case, the corresponding subgroup is called a pj -complement of G, it has order rj o(G)/pj . For example, a group of order 30 (soluble by Theorem 11.7) will have a 2-complement, that is, a subgroup of order 15; a 3-complement, a subgroup of order 10; and a 5-complement, a subgroup of order 6. One consequence of Hall’s First Theorem is: For all soluble groups G and for all primes p dividing o(G), G has a p-complement. Hall’s Second Theorem gives the converse.

Theorem 11.14 (Hall’s Second Theorem) Suppose G is a ﬁnite group. If G has a p-complement for each prime p dividing o(G), then it is soluble.

The proof of this result relies on Burnside’s pr qs -theorem which we shall prove in Web Section 14.2. It states that if o(G) = pr qs , and p and q are prime, then G is soluble; via the Sylow theory, this is Hall’s Second Theorem in the case when o(G) has exactly two distinct prime factors. As with the second derivation of Hall’s First Theorem above, the proof given below is quite long, but it only uses methods applied previously; the main one being consideration of the ‘smallest counter- example’.

Proof Again we use induction on o(G). Suppose the result is false and G1 is a counter-example of smallest order. So G1 has p-complements for all p dividing o(G1), G1 is not soluble, and the result holds for all groups whose orders are less than o(G1). 240 11 Solubility

We show ﬁrst that G1 is simple. Suppose not, then G1 has a proper non-neutral normal subgroup K.Ifp | o(G1), then by hypothesis G1 has r r a p-complement J of order o(G1)/p where p is the largest power of p dividing o(G1).NowJ ∩ K is a p-complement of K, and JK/K is a p-complement of G1/K (by Theorem 5.8, o(JK/K) = o(J/J ∩ K) | o(J)). This holds for all p-complements J , that is for all primes p dividing o(G). Now by the inductive hypothesis both K and G1/K are soluble, and so G is soluble by Theorem 11.4. We complete the proof by constructing a proper normal subgroup of G1 which, of course, is impossible because we have just shown that we may assume G1 is simple. This then shows that our main assumption—that the counter-example G1 exists—is false and the theorem follows. Suppose

= r1 ··· rn = o(G1) p1 pn where n>2 and ri > 0,i 1,...,n. We may assume that n>2 by Theorem 11.6 and Burnside’s pr qs -theorem. For i = 1,...,n,letHi be a pi -complement of G1 as given by the hypothesis, then

[ : ]= ri = ri G1 Hi pi and o(Hi) o(G1)/pi .

Let L = H3 ∩···∩Hn, by Theorem 5.8 this gives

[ : ]= r3 ··· rn = r1 r2 G1 L p3 pn hence o(L) p1 p2 . Now Burnside’s pr qs -theorem asserts that L is soluble, and by Theo- rem 11.11 it has a minimal normal subgroup K which is an (elementary Abelian) p-group for some p dividing o(L). Relabelling if necessary, we may assume that p = p1, and so K is a p1-group. Applying Theorem 5.8 again,

[ : ∩ ]= r2 ··· rn ∩ = r1 G L H2 p2 pn and so o(L H2) p1 ,

that is, L ∩ H2 is a Sylow p1-subgroup of L. Hence, by Problem 6.10(iii),

K ≤ L ∩ H2, and so K ≤ H2. (11.5)

∩ = r2 Also using the same argument, we have o(L H1) p2 , and so

G = (L ∩ H1)H2.

= r1 r3 ··· rn ∈ = ∈ (Note o(H2) p1 p3 pn .) Therefore, if g G1, then g lh where l L ∩ H1 and h ∈ H2. Further, if k ∈ K,wehave

− − − − − g 1kg = h 1l 1klh = h 1yh where y = l 1kl ∈ K as K L. 11.2 Hall’s Theorems and Solubility Conditions 241

This gives by (11.5)

−1 −1 g kg = h yh ∈ H2 (11.6)

for g ∈ G1 and k ∈ K. Finally, let M be the normal closure of K in G;see Problem 2.25.WehaveK ≤ M G1 and M ≤ H2

Example Let H be the group given by

H = a,b | a5 = b6 = e, b−1ab = a4 , a Frobenius group, see Theorem 6.18. The order of this group is 30, and it contains subgroups (p-complements for p = 2, 3 and 5): b of order 6, a,b3 of order 2 10 (D5), and ab of order 15 (C15). Hence the conditions of Hall’s Second Theorem are satisﬁed and so H is soluble. This fact, of course, also follows from Theorem 11.7.

Solubility Conditions

We end this section by listing some solubility conditions. By definition, G is soluble if it has a subnormal series with Abelian factors. By Theorems 11.2 and 11.3,itis sufficient for the factors themselves to be soluble. For a finite group, each of the Conditions 1 to 10 below is equivalent to solubility.

Condition 1 For some positive integer k,thekth derived subgroup G(k) satisﬁes G(k) =e ; see Theorem 11.10.

Condition 2 All factors of all composition series are cyclic of prime order; see The- orem 9.3. This is an easy consequence of the deﬁnitions.

Condition 3 There exists a normal series all of whose factors are p-groups for various primes p. This is also an easy consequence of the deﬁnitions; see Problem 11.5 for this and the previous condition.

Condition 4 Hall subgroups exist for orders m where (m, o(G)/m) = 1 and m | o(G); see Theorems 11.12 and 11.14 above.

Condition 5 All ‘chief factors’ are elementary Abelian (Problem 9.15). 242 11 Solubility

Condition 6 (Sylow systems) Given a group G,letp1,...,pk be a list of the distinct prime factors of o(G). Suppose for 1 ≤ i ≤ k we choose a single Sylow pi -subgroup Pi , and this set P, say, has the property

PiPj = Pj Pi for 1 ≤ i

Then P is called a Sylow system for G. Using his two major theorems, Hall showed that a group G has a Sylow system if and only if it is soluble. Note that some authors give a slightly different, but equivalent, deﬁnition using Hall subgroups rather than Sylow subgroups.

Two examples First, consider the group H given in the example on page 241. It has order 30, and is soluble by Theorem 11.7, or see the quoted example. One Sylow 3 2 system for H is deﬁned by {P1 =b , P2 =b , P3 =a }, for it is easy to see that PiPj = Pj Pi for 1 ≤ i, j ≤ 3asP2 and P3 are both normal subgroups of H . Second, consider A5 which of course is not soluble. Its Sylow 3-subgroups are cyclic generated by 3-cycles, an example is B =(1, 2, 3) , and its Sylow 5-subgroups are also cyclic generated by 5-cycles, C =(1, 2, 3, 4, 5) is an example. We have BC = CB with a similar result for other choices of the Sylow 3- and 5-subgroups, so conﬁrming Hall’s result in this case.

Condition 7IfG is ﬁnite, it is not possible to ﬁnd a,b,c ∈ G satisfying ab = c and o(a),o(b) = o(b),o(c) = o(c),o(a) = 1, (11.7) when a,b and c are all non-neutral.

This number-theoretic condition was ﬁrst studied by Hall (1937), the equivalence was proved by Thompson (1968). There is a connection with the Burnside’s pr qs -theorem which was used in the proof of Theorem 11.14, for if the condition applies then the order of the group must involve at least three primes. For example, consider S5.Ifa = (1, 2)(3, 4), b = (2, 3, 5) and c = (1, 3, 4, 5, 2), it is easy to see that a,b,c ∈ S5, ab = c, o(a) = 2, o(b) = 3 and o(c) = 5, hence a,b and c satisfy (11.7), and by Theorem 11.8 the group S5 is not soluble.

Condition 8IfG is ﬁnite and every pair of elements of G generate a soluble group.

This equivalence was also established by Thompson in his 1968 paper. For example, we can use this condition to reprove the insolubility of S5 by noting that its insoluble subgroup A5 is generated by two of its cycles, for instance, by (1, 2, 3) and (3, 4, 5).

Condition 9 This equivalence is partly due to Galois and is not quite exact. Galois showed that if G is soluble and H is a maximal subgroup, then [G : H ]=pn for some prime p and positive integer n. 11.2 Hall’s Theorems and Solubility Conditions 243

Using Hall’s theorems, it is easy to show that if G is soluble and p | o(G), then there exists a maximal subgroup J which satisﬁes: [G : J ]=pm for some positive integer m. For a ‘converse’ we have: If for all non-neutral subgroups J of G, there exists a subgroup L of J with the property: [J : L]=pk for some positive integer k, and this applies for all primes p dividing o(J), then G is soluble. For a further discussion of this condition see Rose (1978), page 279. In Prob- lem 11.9, we show that there exists a non-soluble group G1 with a subgroup with index a power of p for every prime p dividing o(G1); so in the condition above we need to consider all subgroups J of G and not just G itself.

Condition 10 The factor group G/(G) is soluble where (G) denotes the Frattini subgroup of G; see also Theorem 10.17.

At the beginning of Section 11.1, we listed some number-theoretic conditions on the group order which imply solubility, none is equivalent to solubility. There are also several conditions of a more group-theoretic nature that imply solubility, some of the simpler ones are as follows:

Condition 11 If all maximal subgroups of a ﬁnite group G have an index which is a prime or the square of a prime, then G is soluble.

In Problem 12.9, we show that L2(7) is simple, hence not soluble, and it has a maximal subgroup index 8. This shows that Condition 11 is best possible.

Condition 12 If all proper subgroups of a ﬁnite group G are nilpotent, then G is soluble. A similar condition applies with ‘nilpotent’ replaced by ‘supersoluble’; see Problem 10.26. Also Thompson has shown that if G has a single nilpotent subgroup which is maximal and has odd order, then G is soluble. The simple group L2(17) has a maximal subgroup of order 16, and so the oddness condition above is essential; see Passman (1968), page 141, for further details.

For example, if o(G) = p2q2 and p q − 1, then all subgroups of G are nilpotent (in fact, Abelian), and it is an example of a soluble group; see Problem 11.4.

Condition 13 There are a number of conditions of the following type. Suppose G is ﬁnite, H,J ≤ G and G = HJ, then G is soluble if certain conditions apply to H and J . We give three of the simpler ones here; see Scott (1964), Chapter 13. (a) H and J are nilpotent and (o(H ), o(J )) = 1; (b) H is nilpotent and J has a cyclic subgroup K with [J : K]=2; (c) H and J are both dihedral or dicyclic.

Some more conditions which imply solubility, but are not equivalent to solubility, are given in the problem section below; see Problems 11.6(i) and (iii), 11.8, 11.10(ii), 11.11, 11.14, and 11.15. 244 11 Solubility

11.3 Problems

Problem 11.1 (i) Complete the following sketch of the proof that the three zeros of a cubic polynomial equation with rational coefficients can be defined using radicals, that is, by using square and cube roots. This was first proved by three early sixteenth century Italian mathematicians—S. del Ferro, N. Fontana (Tartaglia) and G. Cardano. The proof was an important step in the development of algebra (and so of science in general) because it forced mathematicians to take complex numbers seriously (R. Feynmann). (a) Using a suitable substitution show that a general cubic polynomial equation with coefficients in Q can be written in the form

f(x)= x3 + ax + b where a,b ∈ Q.

(b) Suppose z is a root of f(x)= 0. Write z = u + v, and so show that

u3 + v3 + (3uv + a)z + b = 0.

(c) Set 3uv + a = 0 to obtain u3 − a3/(3u)3 + b = 0. Solve this for u3, and hence obtain radical expressions for the zeros of the equation f(x)= 0. (ii) In 1545, L. Ferrari showed that quartic polynomial equations also have radical solutions. You are asked to complete the following sketch proof which was given by R. Descartes in 1637. (a) As above show that a general quartic polynomial equation with coefﬁcients in Q can be written as g(x) = x4 + ax2 + bx + c, where we may assume that b = 0 (for otherwise we have a quadratic in x2). (b) Factorising g(x) as a product of two quadratics x4 + ax2 + bx + c = x2 + rx + s x2 − rx + t ,

show that 2s = r2 + a − b/r and 2t = r2 + a + b/r. Deduce r6 + 2ar4 + a2 − 4c r2 − b2 = 0,

and so complete the proof using part (i).

Problem 11.2 (i) Show that the following groups are soluble (a) Dn,(b)Qn,(c)S4. (ii) Give two proofs of the result: If both G and H are soluble, then so is G × H .

Problem 11.3 (i) Using Problem 3.19, or otherwise, show that SL2(5) is not soluble. (ii) Show that if G is not soluble and o(G) ≤ 200, then o(G) = 60, 120, 168, or 180, and give examples in each case. (Hint. The only simple groups satisfying the inequality above have orders 60 or 168, this will be proved in the problem section of Chapter 12.) 11.3 Problems 245

Problem 11.4 Apply the Sylow theory to show that each group G of order pnq or p2q2 where p and q are primes, is soluble using the following method. In the second case, suppose p

Problem 11.5 (i) Show that a refinement of a soluble series is also soluble series. (ii) Must a soluble group with a composition series be finite? (iii) Prove that a finite group is soluble if and only if all factors of all of its composition series are cyclic of prime order. (iv) Finally, prove that a finite group is soluble if and only if it has a normal series all of whose factors are p-groups for various primes p.

Problem 11.6 (i) Using character theory (Web Chapter 13), Burnside proved that if a group G has a conjugacy class of order pr (p prime and r>0), then G is not simple. Use this to prove that if o(G) = pr qs where p and q are prime, and r and s are non-negative integers, then G is soluble. (Hint. Use Theorem 5.19.) (ii) Feit and Thompson (1963) proved that every group of odd order is soluble, show that this statement is equivalent to: ‘every non-Abelian ﬁnite simple group has even order’. (iii) A group is called ‘metacyclic’ (Theorem 6.18) if all of its Sylow subgroups are cyclic. Show that all metacyclic groups are soluble.

Problem 11.7 Suppose o(G) > 1. Show that (i) if G is soluble it has a non-neutral normal Abelian subgroup, and (ii) if G is not soluble it has a non-neutral normal perfect subgroup.

Problem 11.8 (i) Give an example of a soluble group G which is not ‘reverse La- grange’. By reverse Lagrange we mean that G has at least one subgroup of order m for all positive divisors m of o(G), see page 101. (ii) Discuss the proposition: ‘if G is not soluble then it is also not reverse La- grange’.

Problem 11.9 (i) Suppose G is soluble, J G and K is minimal subject to the conditions J

Problem 11.10 (i) Suppose H and J are soluble normal subgroups of a group G. Show that HJ is also a soluble normal subgroup of G, and so deduce that if HJ = G then G is soluble. (ii) Let S(G) be the product of all soluble normal subgroups of G where o(G) < ∞. Show that S(G) is the maximal soluble normal subgroup of G. (iii) Can a maximal normal subgroup of G be larger than S(G)?

Problem 11.11 Find a non-soluble group H which has a subgroup of index p for all p dividing o(H). One example is a direct product of a simple group and a suitably chosen cyclic group.

Problem 11.12 Suppose G is a soluble group, and G(n)/G(n+1) is cyclic for n = 1, 2,.... Using the following method, show that G(2) = G(3) =e .First, use Theorem 11.10 to show that G(3) =e and so G(2) is cyclic, then use the N/C-theorem (Theorem 5.26)toshowG(2) ≤ Z(G ), and apply Problem 4.16(ii).

Problem 11.13 Using the previous problem and Burnside’s Normal Complement Theorem (Theorem 6.17) show that if G is ﬁnite and all of its Sylow subgroups are cyclic, then both G and G/G are also cyclic. This is the starting point for the proof of Theorem 6.18.

Problem 11.14 (i) A theorem of O. Schmidt states: If every maximal subgroup of a ﬁnite group G is nilpotent, then it is soluble. Give a proof of this result using the following method.

(a) Let G1 be a counter-example of smallest order. Show that G1 is simple. (b) Prove that there exist maximal subgroups H1 and H2 of G1 with the property H1 ∩ H2 > e . If not, take conjugates and count elements. (c) Choose H1 and H2 as in (b) so that J = H1 ∩ H2 has the largest possible order, then show that J cannot exist, and so deduce Schmidt’s result. (ii) Use (i) to show that if all proper subgroups of a ﬁnite group G are Abelian, then (a) at least one is normal, and (b) G =e .

Problem 11.15 If G has Abelian subgroups H and J with the property G = HJ, prove that G is soluble. One method uses Problem 2.17 several times to show that G is Abelian.

Problem 11.16 For the groups A5, A6 and GL3(2), see Chapter 12, determine for which prime divisors p of their orders they do not possess p-complements.

Problem 11.17 Given a integer n and primes p and q which satisfy p5, n it is prime, and H Sn−1. 11.3 Problems 247

Problem 11.18 (Properties of Hall Subgroups) Derive the following properties of Hall subgroups, see Deﬁnition 11.13. (i) Suppose G is a ﬁnite soluble group and π is a set of prime numbers. If H is a maximal π-subgroup of G (so there is no π-subgroup of G lying strictly between H and G), then H is a Hall π-subgroup of G. (ii) Let π1 ={3, 5} and consider the group S5. Show that both C3 and C5 are isomorphic to maximal π1-subgroups of S5. Hence maximal π-subgroups of a non-soluble group need not be isomorphic. (iii) The symmetric group S5 has maximal {2, 5}-subgroups which are not Hall {2, 5}-subgroups. (iv) Consider the group GL3(2); see Sections 3.3 and 12.2. This group has Hall {2, 3}-subgroups which are not conjugate. (v) Finally, show that L2(11) (Problem 12.9) has non-isomorphic Hall {2, 3}- subgroups. (Hint. These subgroups have order 12.) Gross (1987) has shown that if 2 ∈/ π, then all Hall π-subgroups are conjugate.

Problem 11.19 Let H beaHallπ-subgroup of G. Derive the following properties. (i) If J ≤ G and o(J) = o(H), then J is a Hall π-subgroup of G. (ii) For g ∈ G, the conjugate subgroup g−1Hg is a Hall π-subgroup of G. (iii) If H ≤ J ≤ G, then H is a Hall π-subgroup of J . (iv) LK/K is a π-subgroup of G/K,ifK G and L is a π-subgroup of G.

Problem 11.20 (i) Give Sylow systems for the groups discussed in Chapter 8. (ii) Why does the group S5 not have a Sylow system? (iii) Show that the solubility of S4 can be deduced from Condition 6 or Condi- tion 7 both of which are given on page 242.

Problem 11.21 (Project—Maximal Subgroups of Soluble Groups) Suppose G is soluble with maximal subgroup H , see also Suzuki (1986), Section 5. Firstly, revisit Problem 11.9(ii) which is due to Galois. Secondly, suppose K = core(H ) =e and let J be a minimal normal subgroup of G. Prove that (a) JH = G and J ∩ H =e . (Hint. Use ﬁrst part.) (b) If H1 maximal, then J ≤ H1 or H1 is conjugate to H in G. (c) J = CG(J ), and it is the unique minimal normal subgroup of G. Lastly, suppose H is another maximal subgroup of G. Show that the following four conditions are equivalent. (d) H and H are conjugate. (e) If G = H L for L ≤ G, then G = HL. (f) G = H H . (g) core(H ) = core(H ). Chapter 12 Simple Groups of Order Less than 10000

The Jordan–Hölder Theorem states that every finite group can be ‘constructed’ from simple groups using extensions. The nature of these extensions is not fully understood (Section 9.2), but a complete list of all finite simple groups is now known—indeed its discovery and development was one of the greatest achievements of twentieth-century mathematics; see the ATLAS (1985), Gorenstein (1982), and Gorenstein et al. (1994). To add to our collections of group examples we shall give an account of the groups in this list whose orders are less than 10000. This may seem a large bound (Appendix D) but, if we exclude Abelian groups, it contains only 16 non-isomorphic groups of three main types (b), (c) and (d) below, and gives a brief ‘snapshot’ of the total picture. The collection of all finite simple groups can be put into four broad categories:

(a) Cyclic groups Cp of prime order p, measured by the size of their order ‘most’ simple groups are of this type, there are 1229 with order less than 10000; (b) Alternating groups An,forn>4, three have order less than 10000; (c) A number of collections of matrix groups defined over finite fields including the linear groups, and a number of other types with similar constructions called the Classical Groups and the Chevalley Groups; there are 16 with order less than 10000 but some isomorphisms between individual groups occur, see below; (d) Twenty-six sporadic groups having a variety of constructions, only one of which has order less than 10000.

As noted above, there is some overlap between these classes; see Problems 6.16, 12.4 and 12.13, Web Section 12.6, and Conway and Sloane (1993). Categories (a) and (b) above have been discussed previously in Chapters 3 and 4 where their simplicity properties were established. Each alternating group An has a presentation of the form | 3 = 2 = An a1,...,an−2 ai (aiaj ) e ,

H.E. Rose, A Course on Finite Groups, 249 Universitext, DOI 10.1007/978-1-84882-889-6_12, © Springer-Verlag London Limited 2009 250 12 Simple Groups of Order Less than 10000 provided n ≥ 4, 1 ≤ i, j ≤ n and i = j. This can be shown to be correct using a similar method to that given in Problem 3.21 for a presentation of Sn.AlsoA5,A6 and A8 have easily defined matrix representations, these are given on pages 260 and 295. In the first section of this chapter, we introduce a new method for constructing groups using so-called Steiner systems. This method is not more important than several others, but we give it because it can be defined with a minimum amount of preliminary material, and it illustrates some of the more complex methods used. The second section introduces the linear groups, see Category (c) above, provides a proof of their simplicity (in all but two cases), and discusses some of their other representations. The third section introduces one class of classical groups—unitary groups, and discusses an example in this class, U3(3), in some detail. The existence of the sporadic groups (Category (d) above) is perhaps the most remarkable single feature of group theory. The first five, the Mathieu groups,were discovered between 1861 and 1873 although their simplicity was established by others two decades later. The remaining 21 were discovered between 1965 and 1980, the first J1 by Z. Janko in 1965. Reasons for this timing include the introduction of electronic computation at this time, and the work of Brauer, Suzuki and others beginning in the 1950s that stimulated mathematicians to take up afresh the challenges of finite group theory. For details on the discovery and early history of the sporadic groups, see Aschbacher (1994) and Solomon (2001). A wide range of methods are used in the construction of simple groups including permutation and matrix theory, certain lattice types (where the groups arise as automorphism groups of these lattices) and game theory, et cetera;1 Steiner systems are one of these. This chapter is more ‘descriptive’ than the others in this book, and many proofs are either sketched or omitted altogether. In some cases, the lack of proof is partly offset by the provision of numerical examples. A whole book would be needed to give all the background and proofs. The properties of the outer automorphism (see pages 81 to 84) and the Schur multiplier groups, and their connections with extensions of the simple groups under discussion, will be treated in Web Section 12.6. The reader is strongly urged to refer to the ATLAS (1985) and the other references for further details. We wish to thank Robert Curtis for advice and helpful suggestions made during the writing of this chapter.

12.1 Steiner Systems

In Chapter 3, we introduced three methods for constructing groups. Although these are the main ones in the theory, a number of other constructions have been inves- tigated, and these need to be considered if all simple groups are to be constructed.

1For more details, see Conway and Sloane (1993). 12.1 Steiner Systems 251

Some occur as ‘automorphisms’ of lattices or ‘games’, and this holds for the constructions to be studied in this section. Further methods are given in the ATLAS (1985), and Conway and Sloane (1993). As noted above the Steiner system method is not more important than several others, but it only requires a minimal amount of preliminary material. Some authors consider that these systems should be named after Kirkman rather than Steiner. T.P. Kirkman was an English mathematician who lived from 1806 to 1895, and J. Steiner was Swiss and lived from 1796 to 1863. We begin with the basic

Deﬁnition 12.1 Let r, s and t be integers satisfying 1

It is not known precisely for which r, s and t Steiner systems exist (but see Prob- lem 12.3). There is no system S(2, 3, 4), for instance, but versions of S(2, 3, 7) do exist as the following example shows.

Example 1 Let X1 ={1, 2, 3, 4, 5, 6, 7} (so t = 7), r = 2, s = 3, and let Z1 = {1, 2, 4}, {1, 3, 7}, {1, 5, 6}, {2, 3, 5}, {2, 6, 7}, {3, 4, 6}, {4, 5, 7} . (12.1)

It is a straightforward exercise to check that every 2-element subset of X1 occurs in exactly one member of Z1, for instance, the pair {1, 5} occurs in the third member of Z1 and no other. This system can be generated as follows: Working modulo 7, start with the triple {1, 2, 4}, that is, the non-zero quadratic residues modulo 7, and apply the map x → x + 1 six times. As an exercise the reader should construct a version of the Steiner system S(2, 3, 9); it has 12 members, see Problem 12.2.

A relatively simple counting argument gives the order of S(r,s,t), provided this system actually exists, as follows.

Theorem 12.2 Using the notation set up in Deﬁnition 12.1, if the Steiner system S(r,s,t)exists, then

t(t − 1) ···(t − (r − 1)) o(Z) = . s(s − 1) ···(s − (r − 1)) 252 12 Simple Groups of Order Less than 10000

Proof Let V denote the collection of all pairs (Y, Z) where Y ∈ Y, that is, Y is an r-element subset of X, and Z is a block in Z. Further, let W denote the subset of V of those pairs (Y, Z) where Y ⊂ Z. We count W in two ways. First, by hypothesis, each Y ∈ Y occurs in exactly one Z ∈ Z,so t o(W) = o(Y) = = t(t − 1) ··· t − (r − 1) r!, r

the number of ways of choosing r elements from a t-element set. Second, W Z s o( ) is the product of o( ), the number of blocks, and r , the number of ways of choosing an r-element subset from an s-element block. Hence s o(W) = o(Z) · = o(Z) · s(s − 1) ··· s − (r − 1) r!. r

The result follows from these two equations by cancelling the term r!.

An extension of this result is given in Problem 12.3. A Steiner system for r, s and t is never unique, for instance, in Example 1 above, the set {3, 2, 4}, {3, 5, 1}, {3, 6, 7}, {2, 5, 6}, {2, 7, 1}, {5, 4, 7}, {4, 6, 1} (12.2) is another version of the Steiner system for the parameters r = 2, s = 3 and t = 7. Compared with (12.1), the elements of the set X ={1,...,7} have been permuted by the 5-cycle (1, 3, 5, 6, 7). But, for example, if we permute the elements of X in (12.1) by the permutation ξ1 = (1, 6, 2)(4, 5, 7) we obtain the Steiner system {6, 1, 5}, {6, 3, 4}, {6, 7, 2}, {1, 3, 7}, {1, 2, 4}, {3, 5, 2}, {5, 7, 4} , that is, we obtain the original system (12.1) again, the only difference being that the order of the sets and the order of the elements in these sets have been changed by ξ1, which is immaterial. We shall see below that the transformation ξ1 is a member of the automorphism group of the Steiner system S(2, 3, 7). These examples suggest that we should make the following

Deﬁnition 12.3 For i = 1, 2, suppose Zi is a Steiner system for Xi . These systems aresaidtobeisomorphic if

(a) there exists a bijection ξ : X1 → X2, and (b) Z ∈ Z1 if and only if Zξ ∈ Z2.

The isomorphism ξ is called an automorphism if X1 = X2 and Z1 = Z2.

In the examples given above, the system (12.1) is isomorphic to (12.2), and the permutation ξ1 is an automorphism of the system Z1. In some cases, if a Steiner system exists for r, s, t , then there is effectively only one such system, that is, they are all isomorphic as in the example above. But in many cases, this is not so, for 12.1 Steiner Systems 253 instance, there are two non-isomorphic systems for the parameters 2, 3, 13 (Prob- lem 12.2) and 80 non-isomorphic systems for the parameters 2, 3, 15. The automorphisms of a Steiner system form a group as we can easily show now.

Theorem 12.4 The collection of automorphisms of the Steiner system S(r,s,t) forms a group under composition which is isomorphic to a subgroup of symmetric group St .

Proof Composition of bijections is closed and associative, and the inverse of a bijection is a bijection (Appendix A). In fact, as all sets are ﬁnite, this also follows from Theorem 2.7(iii). The identity map ι acts as the neutral element. Note that Condition (b) in Deﬁnition 12.1 applies as Z1 = Z2 in all cases. Finally, as ξ is a bijection on X its images are permutations of X, and hence they are elements of St .

The group given by Theorem 12.4 is denoted by Aut(S(r,s,t)). We shall illustrate this result by returning to Example 1. In this case, Aut(S(2, 3, 7)) GL3(2), see Section 12.2,soS(2, 3, 7) has 168 automorphisms. The system S(2, 3, 7), and some other Steiner systems, can be treated as a kind of ‘finite geometry’, that is, as a geometry defined over a finite field. Working over the 2-element field F2 and in ‘dimension 2’, the ‘plane’ of this geometry consists of seven ‘points’ labelled A,B,...,G and seven ‘lines’ where each line consists of exactly three points. This is illustrated in the diagram below which is called the Fano Plane.

In this diagram the strokes, A to B to D, for example, are not part of the geometry, but are included to indicate which sets of three points form lines, so {A,B,D} is one of the seven lines. If we refer now to Example 1 and (12.1), we see that the seven 254 12 Simple Groups of Order Less than 10000 members of S(2, 3, 7) correspond to the seven lines in the diagram above if we set A → 1,B → 2,C → 3,D→ 4,E→ 5,F → 6 and G → 7. In this geometry, as in most others, two distinct points determine a line; so in S(2, 3, 7) there is a unique line (triple of points) passing through each pair of points as given by the properties of this Steiner system. To be more precise, the geometry given in this diagram is a kind of 2-dimensional ‘projective geometry’ deﬁned over F2. Note also that there is no notion of ‘end-point’, so it might be better to treat each of the seven strokes as closed loops. We claimed above that Aut S(2, 3, 7) GL3(2) L2(7).

We will not prove this fact as it would take too long to set up the basic geometric properties needed, the following sketch will suffice; further details can be found in Rotman (1994), Chapter 9, for example. The ‘affine’ part of our geometries can be treated as the geometry of the underlying space of a vector space V , and the linear maps (automorphisms) correspond to the elements of the general linear group GL(V ) represented by the collection of m × m non-singular matrices where m = dim V , they map points to points, and lines to lines. This work can easily be extended to ‘projective’ geometry. Effectively this can be done by adding one extra coordinate, and then the seven points in the diagram on page 253 have the projective coordinates: A = (0 : 0 : 1), B = (0 : 1 : 1), C = (1 : 1 : 1), D = (0 : 1 : 0), E = (1 : 0 : 0), F = (1 : 0 : 1), and G = (1 : 1 : 0) working modulo 2; note that this choice of coordinates is not unique. Hence in our example there is a correspondence between the automorphisms of S(2, 3, 7) and the non- singular 3-dimensional linear maps defined over the 2-element field, that is, GL3(2). By Problem 12.7 , this group is isomorphic to L2(7), in the next section we prove that both are simple and they have the same order. A number of methods are used to construct Steiner systems. Due to space considerations, we shall not be able to give the basic theory or many proofs, but practical procedures will be given for some individual systems. For more details, the reader should consult Cameron and van Lint (1991) or Hughes and Piper (1985).

12.2 Linear Groups

This section introduces the linear groups sometimes called the projective special linear groups. We begin with a brief discussion of ﬁnite ﬁelds, they underlie all the matrices and groups discussed.

Finite Fields

m The finite field Fq has q elements where q = p . These fields were discovered by Galois and as a consequence are sometimes called Galois fields. Their basic properties are as follows: 12.2 Linear Groups 255

(i) Fq is a field, that is, the standard operations of addition, subtraction, multiplication and division apply without restriction, except that division by zero is not allowed. Divisors are constructed using the Euclidean Algorithm (Theo- rem B.2). m (ii) o(Fq ) = q = p where p is a prime and m is a positive integer. In fact, Fq can be treated as a vector space of dimension m over the p-element field Fp with a vector multiplication. So in particular Fq has characteristic p, that is, we work ‘modulo p’. (iii) For each prime p and positive integer m, there is a unique (up to isomorphism) field with pm = q elements. (iv) The multiplicative group of the field Fq is cyclic; see Problem 7.5(iii). (v) In all fields, and so in particular in Fq , a one-variable polynomial equation of degree n cannot have more than n roots, see Theorem B.13 in Appendix B.

For example, consider the ﬁeld F4. Its elements can be written in the form 0, 1, c and c + 1, where c2 = c + 1 and c3 = 1. We have c−1 = c + 1, (c + 1)−1 = c, and c can be taken as the generator of the multiplicative group of the ﬁeld; see (iv) above. For further details, see Problem 12.1 and a standard text on modern algebra, for example, Herstein (1964).

2-Dimensional Linear Groups

In this subsection and the next, we discuss some simple matrix groups. The ﬁrst class is deﬁned by

Deﬁnition 12.5 Given the underlying ﬁeld Fq and the dimension n ≥ 1, the linear group (or sometimes the projective special linear group) Ln(q) is formed by factoring the special linear group SLn(q) (page 53) by its centre.

Some authors write PSLn(q) for Ln(q). We shall show that these groups are simple, except when n = 2, and q = 2 or 3. The proofs follow similar lines to those used to show the simplicity of An for n>4, and given in Chapter 3. In particular, the role played by the 3-cycles in those proofs is played here by a set of matrices called transvections. These are given by

Deﬁnition 12.6 The n × n transvection Ei,j (r) is the matrix formed from the n × n identity matrix In by replacing 0 with r at the (i, j)th place where r = 0 and i = j.

Notes (a) The identity matrix is not a transvection, that is, the inequalities in the above deﬁnition are essential. (b) The determinant of a transvection equals 1. (c) The inverse of a transvection is another transvection. The reader should check these facts. 256 12 Simple Groups of Order Less than 10000

We begin by determining the centre of SLn(F ) for an arbitrary ﬁeld F , this will lead to a formula for the order of Ln(q).

Theorem 12.7 (i) The centre Z of SLn(F ) is the set of matrices aIn where In is the n × n identity matrix, a ∈ F , and an = 1. n(n−1)/2 n 2 (ii) The order of Ln(q) equals q (q −1) ···(q −1)/d where d = (n, q −1), the GCD of n and q − 1.

n Proof (i) Note ﬁrst that each matrix of the form aIn, where a = 1, belongs to Z. For the converse, let A = (aij ) ∈ SLn(F ), and consider E1,2(r).Asr = 0 we have ⎛ ⎞ a11 + ra21 a12 + ra22 ··· a1n + ra2n ⎜ ⎟ ⎜ a21 a22 ··· a2n ⎟ E1,2(r)A = ⎜ . . . . ⎟ , ⎝ . . .. . ⎠ an1 an2 ··· ann

and ⎛ ⎞ a11 ra11 + a12 a13 ··· a1n ⎜ ⎟ ⎜ a21 ra21 + a22 a23 ··· a2n ⎟ AE1,2(r) = ⎜ . . . . . ⎟ . ⎝ ...... ⎠ an1 ran1 + an2 an3 ··· ann

If these two matrices are equal then, as r = 0, a21 = a23 = ··· = a2n = 0, a21 = a31 =···=an1 = 0, and a12 + ra22 = ra11 + a12, which gives a11 = a22. Applying exactly similar arguments using E1,j (r) for 2 ≤ j ≤ n,itfol- lows that if A commutes with all transvections, then all off-diagonal entries in A are zero, and each main diagonal entry equals a11 = a, say. This proves (i). (ii) Use (i), Theorems 3.15 and 4.11, and Problem 7.5(iii) as d = (n, q − 1) n is the number of solutions of the equation a = 1inFq .

The matrices aIn in the proof above are called scalar matrices.Forn = 2, the equation given in the first part of the theorem above, that is, a2 = 1, has solutions 1 and −1 and none other if the characteristic of F is larger than 2, and the single solution a = 1 if the characteristic is 2, as 1 =−1 in a field of this type. For n = 3, the equation a3 = 1 has the single solution a = 1 if the field characteristic is 3 or is congruent to 2 modulo 3, and it has 3 solutions if the characteristic is congruent to 1 modulo 3; see Problem 12.1. We prove the main simplicity result for Ln(q) in two stages: n = 2, and n>2. The proofs given below have a similar structure, the first uses a minimum amount of linear algebra whilst the second relies on the theory of echelon and rational canon- 12.2 Linear Groups 257 ical forms.2 For the first stage, there are two preliminary lemmas which determine the crucial role played by the transvections. The first is

Lemma 12.8 The group SL2(q) is generated by its transvections.

This group also has a 2-generator matrix representation, see Problem 12.5. = ab ∈ = ∈ F \{ } Proof Let A cd SL2(q) and c 0. We have for r, s, t q 0 , 1 r ab 1 s 10 01 cd 01 t 1 a + cr + (as + b)t + (cs + d)rt as + b + (cs + d)r = . c + (cs + d)t cs + d

As c = 0 we can choose s to satisfy cs + d = 1, then this matrix equals a + (as + b)t + (c + t)r as + b + r . c + t 1

If we choose r and t to satisfy as + b + r = 0 and c + t = 0, respectively, then × this matrix equals the 2 2 identity matrix as the determinant is 1. Therefore, − − r, s t 1 x 1 = 1 x with these choices of and , and as 01 01 ,wehave 1 −r 10 1 −s A = , 01 −t 1 01

a product of transvections, s and t are non-zero as c is non-zero. The entry r may be zero in which case A is a product of two transvections. Secondly, if c = 0 then d = 0 as the matrix A is non-singular, and we can argue as follows. 01= Using the argument above, we have −10 D is a product of transvections (the lower left-hand entry is non-zero), then post-multiplying the given matrix by D will provide a matrix whose lower left-hand entry is non-zero and we can return to the ﬁrst case.

The second preliminary lemma shows the connection between transvections and normal subgroups; note the similarity to Lemma 3.13.

Lemma 12.9 Suppose q>3. If K SL2(q) and K contains a transvection, then K = SL2(q).

2An extension of these results known as ‘Iwasawa’s Lemma’ (see, for example, Cameron’s notes at www.maths.qmul.ac.uk/~pjc/class_gps/ or Web Section 5.4) can be used to establish the simplicity of a wide range of matrix groups including Ln(q). 258 12 Simple Groups of Order Less than 10000

Proof By Lemma 12.8, it sufﬁces to show that K contains all transvections. As K is normal, conjugates of elements of K belong to K. Using this fact, 1 r ab ∈ SL (q) suppose 01 is the given transvection, and cd 2 , then d −b 1 r ab 1 + cdr d2r = ∈ K. −ca 01 cd −c2r 1 − cdr

Hence by setting c = 0, and then d = 0, we see that 1 t2r 10 , ∈ K, 01 −t2r 1

for all t ∈ Fq . This gives, as K is also closed under inverses and products, 1 (t2 − u2)r 10 , ∈ K, 01 (u2 − t2)r 1

for all t,u ∈ Fq . There are now two cases to consider: The characteristic of the underlying ﬁeld is (a) not equal to 2, or (b) exactly 2.

Case (a): Characteristic larger than 2.

2 2 We have a = ((a + 1)/2) − ((a − 1)/2) for all a in Fq . This provides suitable values for t and u above, and so all transvections belong to H .

Case (b): Characteristic equals 2. Suppose A = 1 r ∈ K, then as above 10, 1 ra2 and 10also 01 −r 1 01 −ra2 1 belong to K for all a ∈ Fq .Nowfora,c ∈ Fq ,wehave 2 − 2 2 2 2 10 1 rc 10= 1 r a c rc −r 1 01 −ra2 1 −r(1 + a2 − r2a2c2) 1 − r2c2

is in K. If we choose r to satisfy a−1 = 1 + rc (note the map r → 1 + rc is a bijection), then 1 + a2 = r2a2c2 (remember the ﬁeld characteristic is 2) and the lower left-hand entry in the matrix above is 0. Hence with this value of c, 1 b if we construct the commutator of this matrix with 01 we obtain (the square brackets denote the commutator) 1 − r2a2c2 rc2 1 b 1 d , = ∈ K, 01− r2c2 01 01

this is valid as K is normal. Also − d = 1 − r2c2 r2c2 1 − a2 b = a 4 − 1 b 12.2 Linear Groups 259

using the value of r chosen above; reader, check. Now as the order of the ﬁeld is a power of 2, we can choose a to satisfy a−4 = 1asq>3. Hence, as the above calculation is valid for all b ∈ Fq , d can take all values in Fq , that is K contains all transvections.

To be precise, the condition q>3 in the lemma above is not necessary because none of the normal subgroups of either SL2(2) or SL2(3) contain a transvection; the reader should check this using Section 8.2. We can now derive the main result in the 2-dimensional case.

Theorem 12.10 If q>3, the group L2(q) is simple.

Proof We prove this result by showing that if K SL2(q) and K properly contains the centre Z, then K contains a transvection. Lemma 12.9 then shows that K = SL2(q), that is, SL2(q) contains no proper normal subgroup larger than Z if q>3. The simplicity of L2(q),forq>3, follows by the Correspon- dence Theorem (Theorem 4.16). Hence it remains to show that the supposed normal subgroup K contains a transvection if q>3. = ab ∈ ∈ Case 1. Suppose ﬁrst B 0 d K and B/Z. = = = =− If a 1, then d 1 as det B 1, and B is a transvection. If a 1 then =− −10 ∈ ≤ d 1, and as 0 −1 Z K,wehave ab −10 1 −b = ∈ K, 0 d 0 −1 01

and again K contains a transvection. In both subcases, b = 0 because B/∈ Z. This case does not apply if the ﬁeld characteristic is 2. Lastly, suppose a = ±1, and so d = ±1. We have, as K SL2(q),

− − 011 ab 1 01 a 0 = ∈ K, −10 0 d −10 bd

and

− 1 −1 1 ab 1 −1 d 0 = ∈ K, 10 0 d 10 d − a − ba

hence a 0 d 0 ad 0 10 = = ∈ K, bd d − a − ba d2 − ad ad d2 − 11

because ad = det B = 1. Therefore, as d = ±1, K contains a transvection. Note that this last case does not require b to be non-zero. 260 12 Simple Groups of Order Less than 10000 = ab ∈ = Case 2. Secondly, suppose B cd K and c 0. We reduce this to Case 1. As K is a normal subgroup we have 1 −t ab 1 t a − ct (a − ct)t + b − dt = ∈ K, 01 cd 01 cct+ d

for all t ∈ Fq . So, if t satisﬁes a = ct (note c = 0), we obtain 0 b − dt ∈ K, (12.3) ca+ d

and c(dt − b) = 1 as this matrix has determinant 1. Further, if r, s ∈ Fq and rs = 1, then

− − r 0 1 0 b − dt 1 r 0 0 b − dt 0 s ca+ d 0 s ca+ d s2 (a + d)(b − dt)(1 − s2) = , (12.4) 0 r2

and this matrix belongs to K. Therefore, we can return to Case 1 provided 2 2 r, s ∈ Fq \0 satisfy r = ±1 = s , and this is possible for all q except q = 5. We can also return to Case 1 if q = 5 and (a + d)(b − dt) 1 − s2 = 0.

We have b − dt = 0 by the note above, and we can choose s to satisfy s2 = 1. Hence ﬁnally we need to consider the case q = 5 and a + d = 0. By (12.3), 0 u ∈ K u = b − dt = −uc = c 0 where 0(as 1), and 1 v 0 u 1 −v 0 −u c2v2 + 1 v = ∈ K, 01 c 0 01 −c 0 c2v 1 v ∈ F v cv = 2 v for all 5. Choose to satisfy 1, then this matrix equals c 1 with c = 0 = v. Hence we can return to the ﬁrst subcase of Case 2. We have constructed a transvection in all cases, so the theorem follows.

The cases q = 2 and 3 are genuine exceptions for L2(2) GL2(2) S3 (Prob- lem 4.2) and L2(3) A4 (Problem 4.4(iv)). In both cases, the groups contain proper non-neutral normal subgroups. We also have

L2(4) L2(5) A5,L2(7) L3(2) and L2(9) A6. 12.2 Linear Groups 261

The ﬁrst of these statements follows from Problem 6.16(ii), the second from Prob- lem 12.7, and for the third see Problem 12.4. At the end of the last section, we showed that GL3(2) is isomorphic to the automorphism group of the Steiner system S(2, 3, 7), thus providing another representation of this group. Further, some presentations have been constructed for these groups. We have p 4 (p+1)/2 2 3 2 L2(p) a,b | a = a ba b = e, (ab) = b , (12.5) provided p is prime and greater than 3; see Problem 12.6. It can also be shown that o(b) = 2 in this presentation. Another presentation is given in Conway and Sloane (1993), page 268; in some ways, it is more natural being related to work presented in Problem 12.8, but it only applies in the case p ≡ 1 (mod 4). Permutation representations for many of these groups have been constructed. Each of the groups L2(q) is isomorphic to a subgroup of Sq+1 (and in some cases Su for suitably chosen u ≤ q)—a result due to Galois; see Problem 12.8 and Ap- pendix E. We consider some subgroups of one of these groups in Problem 12.9.

Groups Ln(q) with n>2

The general case for Ln(q) where n>2 will be treated now. In the main proof below, we make use of some standard results from linear algebra, in particular the reduced echelon and rational canonical forms of a matrix; see, for example, Halmos (1974a)orRose(2002). The structure of the proof is similar to that for the case n = 2, and so we begin with

Lemma 12.11 The group SLn(q) is generated by its transvections.

Proof Use elementary row operations, the details are left as an exercise for the reader; see Problem 12.4.

Next we consider conjugation.

Lemma 12.12 Two transvections belonging to SLn(q) are conjugate in SLn(q), provided n>2.

Proof Case 1: The transvections are Ei,j (r) and Ei,j (s) for r = s, that is, the ‘new entries’ r and s are in the same position in each transvection. −1 Let t = r s and let D1 be the n × n diagonal matrix with t at the (j, j)th place, t−1 at the (k, k)th place, and 1 at all other diagonal places where k is least such that i = k = j. Clearly, D1 ∈ SLn(q) and

−1 = D1 Ei,j (r)D1 Ei,j (s). 262 12 Simple Groups of Order Less than 10000

Case 2: The transvections are Ei,k(r) and Ej,k(r) where i = j, that is, the ‘new entries’ have the same value and horizontal position but are in different vertical positions. Let D2 be In with the following four changes: (i, i)th place has entry 0, (i, j)th place has entry 1, (j, i)th place has entry −1, and (j, j)th place has entry 0. Clearly, D2 ∈ SLn(q) and −1 = D2 Ei,k(r)D2 Ej,k(r). The remaining cases can be dealt with similarly, we leave them as exercises for the reader.

Every non-singular square matrix deﬁned over a ﬁeld F is similar3 to a matrix in rational canonical form C as follows: C is a diagonal block matrix and each block has the form ⎛ ⎞ 010··· 00 ⎜ ⎟ ⎜ 001··· 00⎟ ⎜ ⎟ ⎜ ...... ⎟ ⎜ ...... ⎟ , ⎝ 000··· 01⎠ c1 c2 c3 ··· ct−1 ct where ci ∈ F and c1 = 0. The blocks are called companion matrices.Ift = 1 then C is the 1 × 1matrix(c1). We can now prove our main result in the n-dimensional case; Theorem 12.10 treated the case n = 2.

Theorem 12.13 If n>2, then Ln(q) is a simple group.

Proof By Lemmas 12.9, 12.11 and 12.12, we need to show that if K SLn(q) and K ≤ Z(SLn(q)) = Z, then K contains a transvection. Let A ∈ K. As noted above, we may assume that A is in rational canonical form (as K is normal); there are three cases to consider: Case 1. A is diagonal (all companion matrices are 1 × 1), and the diagonal entries are non-zero and not all equal (for otherwise A ∈ Z). Case 2. All companion matrices of A are 2 × 2or1× 1, and at least one of the ﬁrst type is present. Case 3. At least one companion matrix of A is r × r where r>2. We shall construct a transvection in each case. Case 1. Suppose the diagonal of A has the form

(a1,...,a1,a2,...,a2,...),

3 The term ‘similar’ means ‘conjugate in GLn(q)’, but using Lemmas 12.11 and 12.12 it can easily be shown that this is equivalent to ‘conjugate in SLn(q)’ in the cases under discussion. 12.2 Linear Groups 263

with r1 copies of a1, r2 copies of a2,..., and ai = aj if i = j.Wehave = − ∈ A,E1,r1+r2 (1) E1,r1+r2 (1 a2/a1) K,

that is K contains a transvection. Case 2. A has the block form BO where B is a 2 × 2 companion matrix. OC 01 = − − × − We may assume that B has the form −1 c and C In 2,then 2 n 2 identity matrix. To see this, we argue as follows. Firstly, by pre-multiplying B = C = by a suitable diagonal matrix we may assume that det det 1, and so 01 B = for some a ∈ Fq . Secondly, by constructing the commutator of −1 a DO A with a matrix of the form , we may assume that C = I − .Welet OIn−2 n 2 − D = 1 1/a , this replaces B by the matrix 0 a where d = 2 + 1/a2. 01 −1/a d Finally, pre-multiplying this by the diagonal block matrix, with 1/a 0 in the 0 a top left-hand corner and In−2 in the bottom right, gives A in the required form with c = 2a + 1/a. Now with A in this form we have A,E1,3(−1) = E1,3(c − 1) + E2,3(1) − In,

and the required transvection is given by E1,3(c − 1) + E2,3(1) − In,E1,2(1) = E1,3(−1) ∈ K.

Case 3. Suppose the ﬁrst companion matrix is r × r where r>2, and its (r, 1)th entry is a1 where a1 = 0 (by deﬁnition of companion matrix). As above we have A,Er,1(−1) = E1,2(1/a1) + Er,1(−1) − In ∈ K,

and E1,2(1/a1) + Er,1(−1) − In,Er,1(−1) = Er,2(1/a1) ∈ K,

another transvection in K. This proves the result.

For n>2, only two of these groups occur in our chosen range: L3(2) with order 4 3 168 and L3(3) with order 5616 = 2 · 3 · 13. The ﬁrst is isomorphic to L2(7), see Problem 12.7 (both are simple and have order 168). The second has a number of distinct representations which we discuss now. First, using a suitable computer package it is a straightforward matter to check that ⎛ ⎞ ⎛ ⎞ 020 010 ⎝ ⎠ ⎝ ⎠ L3(3) is generated by 110 and 001, (12.6) 001 100 over F3 where the ﬁrst matrix has order 6 and the second has order 3; there are, of course, many other choices. Further, this group has the presentation 264 12 Simple Groups of Order Less than 10000 6 3 4 2 4 3 3 2 2 2 L3(3) a,b | a = b = (ab) = a b = a b = a , ba b = e . (12.7)

If we map a in this presentation to the ﬁrst matrix in (12.6) and b to the second, then the equations in the presentation follow, and so it is not difﬁcult to show that (12.7) is a correct presentation for L3(3).Thirdly,wehave L3(3) Aut S(2, 4, 13) .

A Steiner system of the form S(2, 4, 13) can be constructed as follows. Noting that {0, 1, 3, 9} is the set of fourth powers modulo 13, apply the map x → x +1 (mod 13) to this set twelve times to obtain the collection of thirteen 4-tuples: {0, 1, 3, 9}, {1, 2, 4, 10}, {2, 3, 5, 11}, {3, 4, 6, 12}, {0, 4, 5, 7}, {1, 5, 6, 8}, {2, 6, 7, 9}, {3, 7, 8, 10}, {4, 8, 9, 11}, {5, 9, 10, 12}, {0, 6, 10, 11}, {1, 7, 11, 12}, {0, 2, 8, 12} .

It is easily checked that every pair of integers, each of which lie between 0 and 12, occur in exactly one of the 4-tuples above. This can be done by showing that the set of differences of elements of the set {0, 1, 3, 9} modulo 13 gives all twelve non- zero integers modulo 13. Hence we have the Steiner system S(2, 4, 13); note that by Theorem 12.4 this system has 13 members. We can associate the plane projective geometry deﬁned over the 3-element ﬁeld with this system, the geometry has 13 points and 13 lines, and each line contains 4 points, see the discussion given on page 253. Hence the set of linear maps for this geometry is precisely L3(3).This Steiner association also shows that the group can be treated as a subgroup of S13 (or to be more precise a (maximal) subgroup of A13), and, for example, L3(3) is isomorphic to the group generated by the permutations:

(1, 4, 6)(2, 3, 7, 10, 11, 8)(9, 13), (1, 2, 3)(4, 5, 6)(7, 8, 9)(10, 11, 12). (12.8) This pair of generators was constructed by the computer package GAP; in the presentation (12.7), if we map the first permutation in (12.8)toa and the second to b, then the relations in (12.7) hold. The maximal subgroups of this group are isomorphic to either S4C2 (C3 × C3) (26 copies in two conjugacy classes), C3 C13 (144 copies in one conjugacy class), or S4 (234 copies also in one conjugacy class). The first of these subgroups is called the Hessian group, it has connections with a particular cubic curve; see Coxeter and Moser (1984), page 98. Two further points. First, Brauer and Wong (see Huppert and Blackburn 1982b, page 343) have shown that if G is a finite non-Abelian simple group, a is an involution in G, and CG(a) GL2(3), then G is isomorphic to L3(3) or M11 (Section 12.4); note the comments on page 104 and Problem 12.10. Second, Thompson (1968) has shown that, apart from some 2-dimensional linear groups and some Suzuki groups (see Web Section 12.6), L3(3) is the only non-Abelian simple group which is minimal, that is, all of its proper subgroups are soluble. 12.3 Unitary Groups 265

12.3 Unitary Groups

A number of simple groups, sometimes known as the classical groups, can be de- F fined over the finite fields q2 using constructions from linear algebra similar to those given in the previous section. They involving various ‘inner’ or ‘Hermitian’ F products, or more general bilinear forms. We work over q2 (where as usual q is a prime power) so that we can use an analogue of complex conjugation over C.Note ∈ F q2 = F that if x q2 , then x x (Problem 7.5), and so every element of q2 can be treated as a q2th power. Hence it is easily checked that the map : → q ∈ F \{ } ξ a a for a q2 0 F 2 has the usual conjugation properties: It is an automorphism of q2 , and ξ is the F = q identity map on q2 . Because of this we usually write a for aξ( a ),alsowelet 0 = 0. F = Suppose V is a vector space of finite dimension n defined over q2 .Ifu ∈ ∈ F = (a1,...,an) V where ai q2 ,fori 1,...,n, the expression

f(u)= a1a1 +···+anan (12.9) is called a non-singular Hermitian form over V .Ann × n matrix A is called unitary 2 m if A ∈ GLn(q ),forsomeq (= p ), and f (uA) = f(u) for all u ∈ V. (12.10)

Note that forms are non-singular. A more general definition for f can be given, but a basis always exists for the underlying vector space V so that f is in the form (12.9). × F The collection of all n n unitary matrices over q2 is denoted by GUn(q), and F is called the n-dimensional general unitary group over q2 . It can be shown that the determinant of a general unitary matrix is a (q + 1)th root of unity, and the subset of GUn(q) of those matrices A with det A = 1 forms a normal subgroup of order o(GUn(q))/(q + 1) which is called the n-dimensional special unitary group F over q2 ; it is denoted by SUn(q). As in the linear case described in Section 12.2,we factor out the scalar matrices (that is, the centre) to obtain simple groups. SUn(q) has d = (q +1,n)scalar matrices. Hence finally we define Un(q) to be the group formed from SUn(q) by factoring out its scalar matrices, it is called the n-dimensional projective special unitary group, or more usually the n-dimensional unitary group, F = = over q2 . These groups are simple except in the following cases: (a) n q 2, see example below; (b) n = 2 and q = 3, in this case U2(3) A4; and (c) n = 3 and q = 2, in this case U3(2) is isomorphic to the semidirect product Q2 (C3 × C3).

Example We consider the case q = n = 2. Let the ﬁeld F4 have elements {0, 1,c,c + 1} where c3 = 1, c2 = c + 1, and we work modulo 2. By direct calculation we see that GU2(2) is generated by the matrices 10 c 0 0 c , , and , 0 c 0 c2 c2 0 266 12 Simple Groups of Order Less than 10000 C × S A = 10 and it has order 18, it is isomorphic to 3 3. For instance, if 0 c and u = (a1,a2) (and so uA = (a1,ca2)), then f (uA) = a1a1 + ca2ca2 = f(u) as cc = c3 = 1, therefore (12.10) applies. Further, the second and third matrices above generate SU2(2) which has order 6, and so is isomorphic to S3 as it is not cyclic. Now we note that d = (3, 2) = 1 in this case, and so U2(2) SU2(2) S3.

The 2-dimensional unitary groups do not give ‘new’ simple groups for we have

U2(q) L2(q), although this does provide a further representation of the simple group L2(q).Itis not difﬁcult to show that 3 3 2 o U3(q) = q q + 1 q − 1 /(q + 1, 3). (12.11)

The only 3-dimensional unitary groups in our chosen range are U3(2) and U3(3). The first of these was discussed above (it is not simple), and so we need to consider 5 3 the second which is isomorphic to SU3(3), and has order 6048 = 2 · 3 · 7, see 2 (12.11). In this case, we work over the field F9 generated by 1 and c where c = c+1 modulo 3, see Problem 12.1. First, it can be shown that the group U3(3) is generated by the matrices ⎛ ⎞ ⎛ ⎞ c + 211 2c + 12c + 11 ⎝ 120⎠ and ⎝ c 20⎠ , 100 100 where the first matrix has order 8 and the second has order 7. Many other pairs of 3 × 3 matrices generate the group, see Problem 12.12. The group can also be represented as a permutation group and as the automorphism group of the Steiner System S(2, 4, 28); no short presentations have been given but one has been constructed for C2 U3(3) as a Coxeter group, viz.: : 2 = 3 = 8 = 8 = = 4 a1,...,a5 ai (aiaj ) (a3a4) (a2a3a4a5) e,a1 (a3a4) for 1 ≤ i

S(2, 4, 28) also has 63 members, and a version of this system can be constructed using C as follows. For each element σ ∈ C, we form a block with four elements of X ﬁxed by σ . After some computations, we obtained the following version of S(2, 4, 28).

Table 12.1 Steiner System S(2, 4, 28) 1,2,3,4 1,5,18,28 1,6,14,23 1,7,15,22 1,8,19,25 1,9,16,27 1,10,17,24 1,11,20,21 1,12,13,26 2,5,14,24 2,6,19,26 2,7,18,27 2,8,15,21 2,9,20,23 2,10,16,25 2,11,13,28 2,12,17,22 3,5,16,22 3,6,17,28 3,7,20,25 3,8,13,23 3,9,19,24 3,10,15,26 3,11,14,27 3,12,18,21 4,5,20,26 4,6,16,21 4,7,13,24 4,8,17,27 4,9,15,28 4,10,18,23 4,11,19,22 4,12,14,25 5,6,7,8 5,9,21,25 5,10,13,19 5,11,15,17 5,12,23,27 6,9,13,18 6,10,22,27 6,11,24,25 6,12,15,20 7,9,14,17 7,10,21,28 7,11,23,26 7,12,16,19 8,9,22,26 8,10,14,20 8,11,16,18 8,12,24,28 9,10,11,12 13,14,21,22 13,15,25,27 13,16,17,20 14,15,18,19 14,16,26,28 15,16,23,24 17,18,25,26 17,19,21,23 18,20,22,24 19,20,27,28 21,24,26,27 22,23,25,28

The reader should check that each pair of elements in X occurs in exactly one member of the above Steiner system. Incidentally, this shows that the group U3(3) is doubly transitive, see Web Section 5.4. This fact can be used to show that the Steiner system given above is a valid one, and that the automorphism group of S(2, 4, 28) is isomorphic to U3(3). Lastly, we note that the maximal subgroups of U3(3) are isomorphic to either C8 ES2(3) (Problem 6.5, 28 copies), L2(7) (36 2 copies), an extension of S4 by C4 (63 copies), or S3 C4 (also 63 copies). Details on this and the related groups can be found in Huppert (1967), pages 233 to 252, where a simplicity proof is given; see also Problem 12.12 and the ATLAS (1985).

12.4 Mathieu Groups

The smallest ‘sporadic’ group is the first Mathieu group M11, it has order 7920. 6 (The others with order less than 10 are M12 and M22, the second and third Mathieu groups with orders 95040 and 443520, respectively, and J1 and J2 = HJ, the first and second Janko groups with orders 175560 and 604800, respectively; see Web Section 12.6.) As with all descriptions in this chapter, we give informally a number of definitions of M11, but only sketch proofs; a good general account is given in Huppert and Blackburn (1982b). In fact, one of the best ways to work with M11 is as the stabiliser of a point in M12, but as we are limiting our discussion to groups of order less than 10000 we cannot take this approach; see Web Section 12.6 In the 1860s and 1870s, É.L. Mathieu (1835–1890) and others were studying k-transitive groups for k>2. A permutation group G defined on a set X ={1,...,n} is k-transitive if given two k-element subsets Y ={y1,...,yk}, 268 12 Simple Groups of Order Less than 10000

Z ={z1,...,zk} of X there is an element α ∈ G with the property yiα = zi for i = 1,...,k. For all n, Sn,Sn+1 and An+2 are n-transitive, so one might expect that other groups with high transitivity would exist, but this is not so. Apart from the symmetric and alternating groups, there are only two that are 4-transitive—M11 and M23; two that are 5-transitive—M12 and M24; and none that are n-transitive if n>5. (The only known proof of these facts requires CFSG, but in the future it should be possible to give a proof without using the classification.) Mathieu discovered M11 and M12 in 1861, and the remaining three in 1873, their simplicity was established in 1896 by Cole and completed by Miller in 1900. This work was not widely accepted until the 1930s when Witt provided simpler and clearer characterisations, the ‘M’ notation is also due to him. Our first definition of M11 is as a subgroup of A11 as follows. Scott (1964), page 286, Robinson (1982), page 202, and Curtis (2007), page 267, give slight vari- ants on this definition; see also Coxeter and Moser (1984), page 99.

Deﬁnition 12.14 Let

α1 = (1, 2, 3)(4, 5, 6)(7, 8, 9), β2 = (1, 2, 4, 6)(3, 9, 5, 8),

α2 = (1, 4, 7)(2, 5, 8)(3, 6, 9), γ1 = (2, 8)(3, 5)(6, 9)(7, 11),

β1 = (1, 8, 4, 9)(2, 5, 6, 3), γ2 = (2, 3)(5, 6)(8, 9)(10, 11), then M11 is the subgroup of A11 generated by the permutations α1,...,γ2 whose cyclic structures are listed above; it is called the First Mathieu Group. (The second Mathieu group M12 can be deﬁned in a similar way, but as a subgroup of A12 and with the extra generator γ3 = (1, 4)(3, 11)(6, 10)(9, 12).)

The permutations α1 and α2 generate an elementary Abelian group A of order 9. Also β1 and β2 generate a copy B of the quaternion group Q2 of order 8. (Note that 2 = 2 = 2 β1 β2 (β1β2) all with order 2.) Further, we have −1 = −1 −1 = β1 α1β1 α2 ,β2 α1β2 α2α1, −1 = −1 = −1 β1 α2β1 α1,β2 α2β2 α2α1 by direct calculation. Hence C = AB is a group of order 72 with A as a normal subgroup. It is clear that B is the subgroup of C of those permutations which ﬁx 7, and B acts transitively on the set T ={1, 2, 3, 4, 5, 6, 8, 9}. Hence C is transitive on T ∪{7}, and so C is 2-transitive on a 9-element set. Now applying methods given in Web Section 5.4 or in Robinson (1982), page 203, it can be shown that if we add the permutation γ1 to C we obtain a 3-transitive group D of order 720, and then if we add the permutation γ2 to D we obtain a 4-transitive group of order 7920 = 11 · 10 · 9 · 8 isomorphic to M11. It is isomorphic because it has also been shown that all (sharply—see Web Section 5.4) 4-transitive groups deﬁned on 9-element sets are isomorphic to M11. In fact, only two sharply 4-transitive simple groups exist (up to isomorphism), they are A6 and M11. 12.4 Mathieu Groups 269

The Mathieu group M11 has some notable maximal subgroups. The subgroup D (with order 720) deﬁned above is maximal, it is isomorphic to an extension of the cyclic group C2 by A6. It can be shown that M11 contains eleven copies of this group in a single conjugacy class. Twelve copies (again in a single conjugacy class) of the linear group L2(11) also occur as maximal subgroups of M11.The remaining maximal subgroups are isomorphic to C2 C (55 copies, see paragraph below Deﬁnition 12.14), S5 (66 copies), and an extension of S4 by C2 (165 copies). We can construct one of the copies of L2(11) as follows. Let

φ = (1, 9, 4, 6, 2, 10, 7, 3, 11, 8, 5), ψ = (1, 8, 4, 9)(2, 5, 6, 3), θ = (1, 8, 9, 4, 7)(2, 6, 5, 10, 3), ν = (1, 2)(3, 5)(4, 6)(7, 10), then φ and ψ generate M11 as given above (Problem 12.14), φ and θ generate a subgroup of order 55, and φ,θ and ν generate a maximal subgroup H of order 660 isomorphic to L2(11). There is nothing special about the permutations φ,ψ,θ and ν, other choices give the same result but note that θ = ψ2φ5ψ2φ4ψ2φ5 and ν = (θφ)−1ψ2(θφ). Further, H is doubly transitive on {1, 2,...,11}. (This can be checked by hand or by using a suitable computer program.) The index of H in M11 is 12, and this fact can be used to provide a 3-transitive representation of M11 on a 12-element set, see Huppert and Blackburn (1982b). On the other hand, M11 is isomorphic to the automorphism group of the Steiner system S(4, 5, 11), and we can use this fact to show that it is 4-transitive on an 11- element set. In Web Section 12.6, we give several methods for constructing the system S(4, 5, 11). This is done by first constructing a version of S(5, 6, 12) (the automorphism group for this system being isomorphic to M12), and then several versions of S(4, 5, 11) can be ‘read-off’. The ATLAS (1985) gives at least ten methods for constructing M12, and most of these can be specialised to provide definitions of M11; the reader should consult this work for further details. The group M11 can also be characterised as the unique (up to isomorphism) simple group of order 7920; for further details, see Huppert and Blackburn (1982b), pages 341 to 365. As a final pair of representations of M11 we give two of its (many) known presentations: 11 5 4 3 4 4 3 2 M11 a,b,c | a = b = c = (ac) = e,b ab = a ,c bc = b a,b,c,d | a2 = b2 = c2 = d2 = (ab)5 = (bc)3 = (bd)4 = (cd)3 = (abdbd)3 = e ;

The ﬁrst of these is related to the second permutation representation given above with a → φ, b → θ, and c → ψ. In Problem 2.28, we showed that a simple group can be generated by its involutions, the second presentation is an example of this fact—in this case only four involutions are needed. As a computer challenge you could try to ﬁnd four involutions in A11 which satisfy the relations given above and so generate another copy of M11 in A11, see Problem 12.14. Further M11 properties can be found in this problem and at Web Section 12.6. 270 12 Simple Groups of Order Less than 10000

12.5 Problems

Problem 12.1 The field F9 of order 9 is used in a number of examples in this chapter. Here you are asked to determine its basic properties. It can be defined as a vector space of dimension 2 (with basis elements 1 and c) over the three element field F3 ={0, 1, 2}, with a vector multiplication which is defined using linearity by

c2 = c + 1, working modulo 3. For 1 ≤ n ≤ 8 and x ∈ F9 determine − cn, x = x3,x+ x and x 1, that is, write each of these entities as linear combinations of 1 and c.

Problem 12.2 (i) Construct Steiner systems S(r,s,t)in the cases where the parameters r, s, t equal (a) 2, 3, 9, (b) 2, 4, 13, (c) 3, 4, 8, and (d) 4, 5, 11. For (d) try to construct the system without using Web Section 12.6. (ii) Find two non-isomorphic Steiner systems for the parameters 2, 3, 13, note this system has 26 members. One method is to find 13 members as discussed on page 251, then find two further 13-member sets each of which include consecutive pairs. (iii) (The Dining Club) A dining club has the following rules: (a) The Chairman, Secretary and Treasurer attend the first dinner. (b) More people are invited to join. (c) Only Club members may attend dinners. (d) At least three members attend dinners. (e) Each member meets all other members at the dinners. (f) No member shall meet another member more than once. (g) For every pair of members d1 and d2, there is at least one dinner that both d1 and d2 attend. Using these rules answer the following questions. 1. What is the size of the club? 2. How many dinners are held? 3. How many members attend each dinner? 4. How many dinners does each member attend? For further details, see O’Hara and Ward (1937), page 17—per J.F. Bowers.

Problem 12.3 You are given a Steiner system S(r,s,t)for the parameters r, s, t . Extend Theorem 12.2 as follows. Suppose a1,...,au ∈ X where o(X) = t and u

Hence show that v is independent of the particular elements ai . Note that as v must be an integer for all u with 1

Problem 12.4 (i) Give a proof of Lemma 12.11. (ii) Show that A6 L2(9). This can be done using a method similar to that given in Problem 12.7, or it can be done using a suitable computer algebra package, one method is as follows. First, show that A6 can be given by the presentation a,b | a5 = b5 = (ab)2 = (a4b)4 = e .

Begin by choosing two 5-cycles in A6 which satisfy these conditions. Then, working in SL2(9) and its cosets modulo the centre, ﬁnd two matrices in this group which also satisfy these conditions. See also Edge (1955).

Problem 12.5 Let p be a prime larger than 3, and let a be a primitive root modulo p (Appendix B). Show that the matrices a 0 p − 11 A = and B = 0 ap−2 p − 10 generate the group SL2(p). What are the orders of the matrices A and B in this group? See also Problem 3.19.

Problem 12.6 (Properties of L3(2)) (i) Over the ﬁeld F2 let ⎛ ⎞ ⎛ ⎞ 010 100 R = ⎝101⎠ and S = ⎝110⎠ . 011 101

By direct calculation, show that 7 4 4 3 2 R = R S = I3 and (RS) = S , and so prove that R,SL3(2) using the presentation of this group given on page 261. Use the fact that L3(2) L2(7), see Problem 12.7 . (ii) Consider the 7-point ‘geometry’ described in the Fano Plane on page 253 related to the Steiner system S(2, 3, 7). Using the coordinates given on page 254, show that if we apply the matrices R and S to this ‘geometry’ we ‘transform the set of points to itself’, and so using the correspondences given above they deﬁne permutations (in A7)oftheset{1, 2,...,7}. Show that the matrix R corresponds to a 7-cycle a and the matrix S corresponds to a 2-cycle × 2-cycle b,see(12.5) on page 261 with p = 7. Hence show that a and b generate an isomorphic copy of L3(2) in A7. (iii) Now using the permutation representation constructed in (ii) show that L3(2) contains subgroups isomorphic to (a) F7,3, and (b) S4. 272 12 Simple Groups of Order Less than 10000

Problem 12.7 (Simple Groups of Order 168) In this question, you are asked to prove that all simple groups of order 168 are isomorphic. These include the groups L3(2) and L2(7) discussed in this chapter and the problem above. One method is as follows. Suppose G is a simple group with o(G) = 168, and P = a is a Sylow 7-subgroup (so o(a) = 7). 7 3 2 4 (i) Let H = NG(P ). Show that H a,b | a = b = e, bab = a for some b ∈ G, that is, H is a non-Abelian group of order 21. Also show that ba = a4b. 3 2 2 (ii) Show that J = NG( b) b,c | b = c = e, cbc = b for some c ∈ G, that 2 is J S3 and cb = b c. (iii) Prove that the set of left cosets of H in G can be written in the form C = H,cH,acH,...,a6cH .

Using this fact, deduce G is generated by a,b and c. (iv) Define a map θ from G to S8 as follows. Number the elements of C from 1 to 8, respectively. The left coset action of g ∈ G permutes the elements of C and so 6 defines an element of S8, that is, {gH,gcH,...,ga cH} is a permutation of C which we denote by gθ. Show that (a) this defines an injective homomorphism from G to S8,(b)aθ = (2, 3, 4, 5, 6, 7, 8), and (c) bθ = (3, 6, 4)(5, 7, 8). (v) Show that cθ can be expressed in three different ways as a product of four 2-cycles σi = (1, 2)(3, ∗)(∗, ∗)(∗, ∗) for i = 1, 2, 3, and if Li = aθ,bθ,σi then Li ≤ A8 again for i = 1, 2, 3. (vi) Conclude that L1 = L2 = L3, and so finally prove that all simple groups of order 168 are isomorphic. We also have: If o(G) = 168 and G has no normal Sylow subgroup, then G L3(2); see Suzuki (1982), page 107.

Problem 12.8 Consider the group L2(11) (of order 660, Deﬁnition 12.5). Give a permutation representation of this group using the following method. Work on the projective line

P ={∞, 0, 1, 2,...,10} deﬁned over F11, ∞ = ab ∈ and treat the symbol in the usual naive way. Given a matrix A cd SL2(11), deﬁne a map θA : P → P by az + b a zθ = for z ∈ P\{∞}, and ∞θ = . A cz + d A c

Show that the map θA permutes the elements of P and so deﬁnes an element in S12. Deduce that L2(11) is isomorphic to a subgroup J of S12. Secondly, let maps α and β on P be given by

zα = z + 1 and zβ =−1/z for z ∈ P.

Prove that these two maps generate J , and so give a permutation representation of L2(11). Finally, show that the presentation given for this group on page 261, with 12.5 Problems 273 p = 11, is a valid one. As an extension to this problem, consider what is needed to establish the general case where 11 is replaced by a general prime p and P has p + 1 elements.

Problem 12.9 Using the permutation representation for L2(11) constructed in Problem 12.8, ﬁnd examples of the maximal subgroups of L2(11). (Hint. Using some results from this and the previous chapters, show ﬁrst that L2(11) has no proper subgroups of order greater than 66, then use a suitable computer algebra package, or try by hand.) Note that this group has an index 11 subgroup isomorphic to A5, and this fact allows us to construct a degree 11 permutation representation for it.

Problem 12.10 The group L3(3) was discussed in Section 12.2 where we proved that it is simple. Show that L3(3) has 117 involutions, and using these involutions construct elements of the group of all possible orders, that is, of orders 1, 2, 3, 4, 6, 8, and 13. Show also that the group can be generated by two elements of order 3, and by three elements of order 2 (cf. Problems 2.28 and 3.20). One way to do this is to begin with the permutation representation (12.8) given on page 264.

Problem 12.11 (i) Continuing the previous problem show that ⎛ ⎞ 001 A = ⎝020⎠ 100 is an involution in L3(3). (ii) Let H denote the centraliser of A in L3(3). Show that o(H) = 48 using the basic properties of the centraliser, and so by Problem 12.10 show that the collection of involutions in L3(3) forms a single conjugacy class in L3(3). (iii) Using Problem 12.6 deduce H GL2(3). Only two simple groups have this property; they are L3(3) and M11, see Huppert and Blackburn (1982b).

Problem 12.12 (i) The group U3(3) contains elements of order 1, 2, 3, 4, 6, 7, 8, and 12. Using the matrices A and B given on page 266 ﬁnd elements of this group having each of these orders. (ii) Prove that U3(3) contains 63 elements of order 2. You may assume that each matrix of order two in the matrix representation of U3(3) given on page 266 has the form ⎛ ⎞ a2 a1 c1 ⎝ ⎠ C = a3 c2 a1 , c3 a3 a2 3 where the overline denotes conjugation (x = x ), ai ∈ F9 and ci ∈{0, 1, 2},fori = 1, 2, 3. (Hint. Use the standard formulas for det C and C−1.) You could also try to prove the assumption. 274 12 Simple Groups of Order Less than 10000

= (iii) If b is an involution in U3(3), show that J CU3(3)(b) C4S4, where C4 J , using the following method. Apply the permutation representation of U3(3) given on page 267, choose an involution b which is represented by a product of twelve disjoint 2-cycles in S28, see the comments on the construction of the Steiner system for this group on page 267. Secondly, ﬁnd an element of order 4 in U3(3) deﬁned on the same symbols as b which commutes with b (note that its square will equal b). Now using (ii) and the fact that the permutation group on the symbols not used in b is isomorphic to S4, construct the remaining elements of J .

This problem is not hard but it will take time to complete and the use of a suitable computer algebra package will help.

Problem 12.13 (Simple Groups of Order 20160) By Theorems 2.15 and 11.7,the simple groups A8, L4(2) and L3(4) all have order 20160, A8 L4(2) (Conway and Sloane 1993), but L3(4) is not isomorphic to either of the other two groups, and in this problem you are asked to prove this fact. There exist infinitely many pairs of non-isomorphic simple groups with the same order, but no triples; this is a consequence of CFSG, see the ATLAS (1985) for details. There are a number of ways to establish the required non-isomorphism property. It can be done by showing that A8 contains elements of order 15 whilst L3(4) does not, or by showing that o(Aut(A8)) = 8! whilst o(Aut(L3(4)) ≥ 3 · 8!/2 (Scott 1964, page 314). We suggest two further methods (i) and (ii) that you can try. (i) Show first that A8 has non-conjugate involutions; then, using the following method, prove that the involutions of L3(4) form a single conjugacy class. We have 2 (a) a non-scalar matrix A in SL3(4) is an involution in L3(4) if and only if A is scalar (that is, diagonal with identical diagonal entries); (b) A2 is scalar if and only if (C−1AC)2 is scalar for all non-singular matrices C ∈ SL3(4); (c) by (b), A can be replaced by a similar matrix in one of the following three rational canonical forms: ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ a1 00 b1 00 01 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ B1 = 0 a2 0 ,B2 = 001,B3 = 00 1, 00a3 0 b2 b3 1 c1 c2 where the entries belong to the field F4, and a1,a2,a3,b1, and b2 are non-zero.

= −1 Show that the matrix A, see (a), can only be of type B2 in (c), b2 b1 and 2 2 b3 = 0, hence the entries b1,b2,b3 of B2 equal either 1, 1, 0; c,c , 0orc ,c,0 where c is a multiplicative generator of the ﬁeld F4. Now consider the squares of these matrices and use Problem 5.21(iii). (ii) The second method is as follows. Firstly, show that both L4(2) and L3(4) have Sylow 2-subgroups consisting entirely of upper triangular matrices (Prob- lem 3.15). (Note that L4(2) SL4(2), and by Theorem 12.7 L3(4) SL3(4)/C3,so in the second case work by identifying the three matrices in each coset.) Secondly, 12.5 Problems 275 show that the centres of both groups also consist of upper triangular matrices, and all matrices in each centre have non-zero entries in their upper right-hand corners. Finally, by considering the orders of these centres, show that the Sylow 2-subgroups of the two groups in question are not isomorphic.

Problem 12.14 (i) Show that the two permutation definitions of M11 given on pages 268 and 269 define the same group. This can be done by a computer search, or by showing that α1,...,γ2 can be defined in terms of φ and ψ, and vice versa. −1 −1 To start with note that α1 equals the conjugate of φψ by θ φ . (ii) Find examples of the Sylow subgroups of M11 for each of the primes dividing its order. (Hint. Using the notation of Section 12.4,wehaveδ = β1β2γ2 has order 8.) (iii) In M11, choose an involution a and show that its centraliser in M11 is isomorphic to an extension of C2 by S4, see Problems 4.4 and 12.9. (iv) Find four involutions in A11 which generate M11 and satisfy the relations given in the second presentation of M11 on page 269. (Hint. One method starts with γ1 and γ2 in Definition 12.14.)

Problem 12.15 (Simplicity of M11) (i) Suppose G is a transitive subgroup of Sp, o(G) = pnt where

n>1,n≡ 1 (mod p), t < p and t is prime.

Apply Problem 6.12(ii) to show that G is simple. Method. Using the quoted problem we can take n = np and t =[NG(P ) : P ] where P is a Sylow p-subgroup of G. Suppose K G and K> e. Show that there exists a Sylow p-subgroup Q of G which is a subgroup of K. Now use Theorem 6.9(ii) and the quoted problem again to show that o(K) = pnps where s | t and s>1, and so deduce K = G; see Chapman (1995). (ii) Use (i) to show that M11 is simple. The same method works for the simplicity of M23 with p = 23 and t = 11. The Mathieu group M23 can be deﬁned using the Steiner system S(4, 7, 23) or as a subgroup of S24, see either Scott (1964), pages 287–289, or the ATLAS (1985), page 71.

A number of the problems above are ‘project-like’, and so no speciﬁc ﬁnal project will be given. Appendices A to E

Appendix A: Set Theory

The elementary ideas and notation of set theory underlie all the work in this book. Here we give a outline of the basic theory and terminology which is sufﬁcient to cover the material presented in the previous chapters. The reader requiring a more thorough treatment should consult Halmos (1974b).

A.1 Sets and Maps

Set theory is based on the following notions: (a) element (or member)—we use lower case letters x,y,...for elements, (b) membership (or belonging)—denoted by ∈, (c) set (or collection)—we use upper case letters X,Y,...for sets. Using logical constructions, all statements can be built up from the basic proposition

x ∈ X, (A.1) that is, the element x is a member of the set X. In an axiomatic development of set theory, these notions are taken as primitives which obey certain axioms and rules. For our purposes, we take an informal approach, assuming that the reader has an intuitive idea of the notion of an element (or member) belonging to a set (or collection), and this is expressed formally in (A.1). We also use the following notation, terminology and constructions. (i) X = Y , X equals Y —every element in the set X also belongs to the set Y , and vice versa. The negation is written X = Y . (ii) ∅,theempty set, is the (unique) set with no elements. (iii) Y ⊆ X, Y is a subset of X—every element in the set Y also belongs to the set X (this is equivalent to X ⊇ Y ,thesetX contains the set Y ). For all X,the

H.E. Rose, A Course on Finite Groups, 277 Universitext, DOI 10.1007/978-1-84882-889-6_13, © Springer-Verlag London Limited 2009 278 Appendices A to E

empty set ∅ and X are subsets of X. We write Z ⊂ X for Z ⊆ X and Z = X, and say that Z is a proper subset of X. (iv) X ∩ Y ,theintersection of X and Y , is the largest set that contains all those elements which belong both to X and to Y (note X ∩ (Y ∩ Z) = (X ∩ Y)∩ Z and X ∩ Y = Y ∩ X). (v) X ∪ Y ,theunion of X and Y , is the smallest set that contains every element of X and every element of Y (note X ∪ (Y ∪ Z) = (X ∪ Y)∪ Z and X ∪ Y = Y ∪ X), see Problem A.1. A union X ∪ Y is called disjoint if X ∩ Y =∅;this is written as X ∪˙ Y . (vi) Y \X,thecomplement of X in Y , is the subset of Y of those elements that do not belong to X (note Y \Y =∅and X\(X\Y)= X ∩ Y ). (vii) If x1,...,xn are elements in some set X, then the (unordered) set containing these elements is denoted by {x1,...,xn}, and so {x} denotes the set with the single element x. We use the notation (x1,...,xn) for the ordered set whose first element is x1, second element is x2,..., and last element is xn. Infinite versions of these constructions are also used. (viii) If P is a property which is either true or false for elements of a set X, then we define the new set Y by Y = x ∈ X : P(x) .

The set Y is the subset of X containing those elements x belonging to X for which P(x)is true. There is another construction which is important in group theory—Cartesian product named after the seventeenth century French mathematician and philosopher René Descartes. Suppose X1,...,Xn is a ﬁnite (or inﬁnite) collection of sets. We form the new set of ordered n-tuples (x1,...,xn) : x1 ∈ X1,...,xn ∈ Xn .

This is denoted by X1 ×···×Xn and called the Cartesian product of Xi ,fori = 1,...,n. Note that the n-tuples are ordered so, for instance, X1 × X2 = X2 × X1, n unless X1 = X2. We write X for X ×···×X with n copies of the set X.

Example Let X, Y and Z be subsets of the set of real numbers R given by

X ={x ∈ R : 0 ≤ x ≤ 1},Y={y ∈ R : 1 ≤ y ≤ 2}, Z ={z ∈ R : 0 ≤ z ≤ 2}.

The following statements are valid; the reader should check them.

(i) X ⊆ Z, X ∪ Y = Z, and X ∩ Y ={1}, (ii) Z\X ={x ∈ R : 1

Relations, Functions and Maps

An n-ary, or n variable, relation R on a set X is by definition a subset of the Carte- sian product Xn; relations are called binary if n = 2, ternary if n = 3,....Sofor example, if X = R, the binary relation x ≤ y, where x,y ∈ R, is defined as the subset of the set of all ordered pairs (x, y) ∈ R×R for which there exists a non-negative z with the property y = x + z. There are a number of important classes of relations. One is the class of equivalence relations; they are defined as follows.

Deﬁnition A.1 Let X be a set and let ∼ be a binary relation on X. (i) The relation ∼ is called an equivalence relation if the following three axioms hold for all x,y,z ∈ X: (a) it is reﬂexive—x ∼ x, (b) it is symmetric—if x ∼ y then y ∼ x, and (c) it is transitive—if x ∼ y and y ∼ z, then x ∼ z. (ii) For z ∈ X,let

Ez ={x ∈ X : x ∼ z}.

The subset Ez ⊆ X is called the equivalence class of z for the relation ∼. We have

(i) x ∈ Ez if and only if x ∼ z, and (ii) X is a disjoint union of its equivalence classes under ∼. ∈ ∈ ∼ ∼ This union is disjoint because if x Ez1 and x Ez2 , then x z1 and x z2, and so ∼ = z1 z2 and Ez1 Ez2 by (b) and (c) above. The simplest example of an equivalence relation is equality; in this case, the equivalence class of x is the singleton set {x}.In fact, equivalence relations are generalisations of equality. See also the congruence subsection on page 285. The second important class of relations contains the maps. We use the words map, mapping, transformation and function interchangeably, see also (vi) below. As noted earlier, see page 68, we write maps ‘on the right’, that is, in algebraic contexts, we write aθ, rather than θ(a), because we read from left to right. Formally, we deﬁne a map as follows:

Deﬁnition A.2 Given sets X and Y ,amap φ : X → Y isarelationonthesetX × Y with the extra property. For each x ∈ X, there exists a unique y ∈ Y such that (x, y) belongs to the relation φ.

Note that a map or function must be single-valued. An equivalent, and more intuitive, definition is to say that: There is some rule,orprocedure, denoted by φ which, given x ∈ X, generates a unique element xφ ∈ Y , and this holds for all x ∈ X. 280 Appendices A to E √ For example, suppose X = Y = R and xφ = x for x ∈ X. As this stands φ is not a function,√ but it is if we always take the positive square-root, that is, if we define xφ = + x. If X = X1 ×···×Xn,themapφ : X → Y is called an n-argument or n-variable function from X to Y , and if (x1,...,xn) ∈ X, the elements x1,...,xn are called its arguments or variables. The basic definitions and facts about maps are as follows. (i) If X and Y are sets and φ : X → Y is a map, then X is called the domain of φ, and Y is called the codomain of φ.Theset{xφ : x ∈ X} is a subset of Y which is called the image,orrange,ofφ, and it is denoted by im φ. With a slight ‘abuse of notation’ we write Zφ for the image of Z under the map φ where Z ⊆ X; that is, Zφ ={y ∈ Y : zφ = y for some z ∈ Z}.Also,xφ is called the image of x; and if zφ = y, then z is called a preimage of y.IfZ ⊆ im φ, then the set of all preimages of elements of Z under the map φ is denoted by Zφ−1. (ii) If φ : X1 → Y1 and ψ : X2 → Y2 are maps, then we say that they are equal, written φ = ψ,ifX1 = X2 and xφ = xψ for all x ∈ X1. Note that we do not require the codomains to be equal, only that the domains and images agree. (iii) A map φ : X → Y is called injective (or sometimes one-to-one)ifxφ = yφ implies x = y, for all x,y ∈ X; that is for each z in the image of φ there exists exactly one preimage w in the domain X with wφ = z. 3 For example, if X = Y = R and xφ1 = x for all x ∈ X, then φ1 is injective because 2 every real number has a unique real cube root, but if xφ2 = x for all x ∈ X, then φ2 is not injective because, for instance, both 2 and −2 are mapped to 4 by φ2. (iv) A map φ : X → Y is called surjective (or sometimes onto) if for all y ∈ Y , there is at least one x ∈ X satisfying xφ = y, or equivalently, Y equals the image of φ.

In the examples given below (iii), φ1 is surjective whilst φ2 is not (for instance, −1 has no preimage). (v) A map is called bijective (or sometimes a one-to-one correspondence)ifitis both injective and surjective. If a bijection exists between two sets (that is, there is a bijective map between them), then they have the same cardinality, or to put this another way, they have the ‘same number’ of elements. This applies in both the ﬁnite and inﬁnite cases. We use o(X) to denote the cardinality (order) of X throughout, that is both for sets and for groups. Some authors use the word size for the cardinality of a set and reserve the word ‘order’ for groups.

The map φ1 given below (iii) is an example of a bijection. As most of the groups discussed in this book are finite, cardinality problems do not arise—the cardinality of a finite set is just the (finite) number of its elements. Problems can arise if the group in question is neither finite nor countably infinite, the interested reader should consult the text by Halmos referred to at the beginning of this Appendix. (vi) Given a non-empty set X,themapι : X → X defined by xι = x for all x ∈ X is called the identity map—each x ∈ X is ‘identified by itself’. It is, of course, a bijection on X. A Set Theory 281

(vii) If Y is a set and X = Y 2, then the map μ : X → Y is called a (binary) operation 2 on Y (but note that it is a ternary relation). For every pair (y1,y2) in Y , there is always a unique y3 in Y with (y1,y2)μ = y3. For instance, if Y = R, we can take μ to be the usual addition operation +. We come now to the important notion of composition of functions which is widely used throughout the book.

Deﬁnition A.3 Given two maps φ : X → Y and ψ : Y → Z, a third map, denoted by φ ◦ ψ : X → Z and called the composition of φ and ψ, is given by

x(φ ◦ ψ)= (xφ)ψ.

Note that (a) as φ maps X → Y and ψ maps Y → Z, the composite φ ◦ ψ maps X → Z, and (b) we read all algebraic formulas and propositions from left to right both here and throughout this book. In our opinion, this is especially helpful when discussing composition.

Theorem A.4 (i) Composition is associative. (ii) If φ : X → Y and ψ : Y → Z are injective (surjective) maps, then φ ◦ ψ : X → Z is also injective (surjective, respectively).

Proof (i) If θ : Z → W , then both (φ ◦ ψ)◦ θ and φ ◦ (ψ ◦ θ) map X to W and we have using Deﬁnition A.3 x (φ ◦ ψ)◦ θ = x(φ ◦ ψ) θ = (xφ)ψ θ, and x φ ◦ (ψ ◦ θ) = (xφ)(ψ ◦ θ)= (xφ)ψ θ.

This holds for all x ∈ X, and so the result follows. (ii) First part. We have, if x(φ ◦ ψ) = y(φ ◦ ψ), then (xφ)ψ = (yφ)ψ which implies, as ψ is injective, that xφ = yφ. This now gives x = y because φ is also injective; the second part is similar.

We usually drop the symbol ‘◦’ and write φψ for φ ◦ ψ. Some examples are given in Problem A.4. If φ : X → Y is a bijection, see above, then we can deﬁne the inverse map φ−1 as follows:

− if y ∈ Y, then let yφ 1 = x where x is given by xφ = y.

Note that φ−1 is a map from Y to X because φ is both injective and surjective. Also if X = Y , then

− − φ ◦ φ 1 = φ 1 ◦ φ = the identity map ι on X. (A.2) 282 Appendices A to E

The first of these equations follows by definition; for the second, we note that if xφ−1 = z and z ∈ X, then zφ = x, and so x φ−1 ◦ φ = xφ−1 φ = zφ = x, and this holds for all x ∈ X. Referring to the example below (iii) on page 280 where −1 φ1 is defined, we see that φ1 is the (unique) real cube root function. Ordered sets play some role in group theory, so we will give a brief outline here.

Definition A.5 (i) A partial order on a set X is a binary relation on X which satisfies the following three axioms. For x,y,z ∈ X, the relation is (a) reflexive—x x, (b) transitive—if x y and y z, then x z, and (c) antisymmetric—if x y and y x, then x = y. (ii) If satisfies (a), (b), (c), and it is (d) symmetric—x y or y x,forx,y ∈ X, then the order is called linear or total. (iii) A total order on X is called a well-order on X if every non-empty subset of X has a least element in X. Induction can be applied to well-ordered sets.

Example 1 If X is a set and P(X) is the set of all subsets of X, and we deﬁne an order on P(X) by: for Y1,Y2 ∈ P(X),letY1 Y2 if and only if Y1 ⊆ Y2, then is a partial order on X.

Example 2 If X is the set of integers Z, then the usual inequality relation ≤ is a linear order on Z, and it is a well-order on the set of non-negative elements of Z.

A.2 Problems

Problem A.1 Let X, Y and Z be sets. Prove (i) X ∪ (Y ∩ Z) = (X ∪ Y)∩ (X ∪ Z), (ii) X ∩ (Y ∪ Z) = (X ∩ Y)∪ (X ∩ Z), (iii) (X\Y)∪ (Y \X) = (X ∪ Y)\(X ∩ Y). Propositions (i) and (ii) are instances of De Morgan’s Laws (named after the nineteenth century English mathematician Augustus De Morgan), they can both be extended ﬁnitely and inﬁnitely, so for example, we have

X ∪ (Y1 ∩···∩Yn) = (X ∪ Y1) ∩···∩(X ∪ Yn).

The expression on either side of the equation in (iii) is called the symmetric difference of X and Y , it is sometimes denoted by XY. A Set Theory 283

Problem A.2 For each of the following sets X with relations ∼ determine which deﬁne systems with equivalence relations, and describe the equivalence classes.

(i) X denotes the collection of all human beings alive at noon today, and (a) x ∼1 y stands for ‘x and y have the same ancestor’ (assume that a person can be his, or her, own ancestor), and (b) x ∼2 y stands for ‘x and y have the same mother’; (ii) X = R, and (a) a ∼1 b stands for ‘x

Problem A.3 Let X and Y be arbitrary sets. (i) Is X ×∅non-empty if X is non-empty? (ii) Show that (X × Y)∩ (Y × X) = ∅ if and only if X ∩ Y = ∅.IfX and Y have n elements in common, how many elements do X × Y and Y × X have in common?

Problem A.4 For the following maps φ : X → Y determine whether they are injective, surjective or bijective. (i) X = Z × (Z\{0}), Y = Q, and (a, b)φ = a/b; (ii) X = Z × Z, Y = Q, and (a, b)φ = a + b; (iii) X = W × Z, Y = Z and (x, y)φ = y—maps of this type are called projections; (iv) X = R, Y = R+, the set of positive real numbers, and aφ = 2a; (v) X = Z, Y = Q and xφ = x/2; (vi) X = R, Y = Z, and aφ =[a]—where the square brackets denote integer part.

Problem A.5 Let Y be a ﬁnite set and let ψ : Y → Y . Show that if ψ is injective, then it is also surjective, and vice versa. Is this still true if Y is inﬁnite?

Problem A.6 Let M beasetwithm elements and N beasetwithn elements. How many functions are there that map M to N, and how many of these are (i) bijections, or (ii) injections, or (iii) surjections.

Problem A.7 Show that the following sets are bijective, that is, they have the same cardinality. (i) X × (Y × Z) and (X × Y)× Z where X, Y and Z are sets; + (ii) Z and GL2(Z); and (iii) Z and Q.

Problem A.8 Prove that φ : X → Y is a bijection if and only if there is another map ψ : Y → X with the properties (a) φ ◦ ψ is the identity map on X, and (b) ψ ◦ φ is the identity map on Y .

Problem A.9 Give examples of operations on a set X which are associative but not commutative, and vice versa. (Hint. Try X as a 2-element set.) 284 Appendices A to E

Appendix B: Number Theory

Some aspects of elementary number theory play a vital role in group theory, especially in the ﬁnite case. In particular, the prime numbers, Euclid’s algorithm, congruences and primitive roots are widely used. Here we give a brief introduction to those aspects of number theory that appear in this book including the four topics listed above; for further details the reader should consult, for example, Rose (1999) where full proofs can be found.

B.1 Divisibility and the Euclidean Algorithm

Number theory begins with the study of the integers Z. This system is a group under addition +. It also has both a linear order ≤ and a multiplication operation, but the non-zero integers do not form a group under this second operation because it has no inverses. This leads to the ﬁrst result

Theorem B.1 (Division Algorithm) Suppose a,b ∈ Z and b = 0. There exist unique integers c and d satisfying

a = bc + d and 0 ≤ d<|b|, where the vertical bars denote the absolute value.

Theorem B.2 (Euclidean Algorithm) Suppose a,b ∈ Z, and at least one of a or b is non-zero. There exists a unique integer c which satisﬁes

c>0,c| a,c | b, and if d | a and d | b, then d | c.

This follows from the Division Algorithm (Theorem B.1) and (iv) above. The unique integer c given by this algorithm is called the greatest common divisor or GCD of a and b, and it is denoted by (a, b);alsoweset(0, 0) = 0. Some authors use HCF (highest common factor) for GCD. The basic properties are given in Prob- lem B.1. We deﬁne the least common multiple or LCM of a and b to equal ab/(a,b) provided both a and b are non-zero. This algorithm is named after the Greek geome- ter Euclid of Alexandria who was working in the third century BC. B Number Theory 285

There are two important corollaries of this algorithm, the ﬁrst is

Theorem B.3 The equation ax + by = n has integer solutions x and y if and only if (a, b) | n.

In Theorem B.2, c = (a, b) is the smallest positive integer of the form ax + by. The second corollary is

Theorem B.4 If a | bc and (a, b) = 1, then a | c.

This is proved by applying Theorem B.3 to the given equation. Using Theorem B.4, we deﬁne the primes and unique factorisation in Z.

Deﬁnition B.5 (i) An integer p>1 is called prime or a prime number if its only positive divisors are 1 and p, otherwise it is called composite. Two integers are called coprime if they have no common positive divisor except 1. (ii) We use the symbol π to denote a collection of primes, that is, a subset of the set of all prime numbers. So a π-number is a positive integer whose prime factors are included in π. Also we write π for the complement of π in the set of all prime numbers.

There are inﬁnitely many primes, this was ﬁrst proved by Euclid. (If p1,...,pk is a list of all primes, consider the expression p1 ···pk + 1.) Unique factorisation in Z now follows directly from Theorem B.4,wehave

Theorem B.6 (Unique Factorisation) Every positive integer has a unique (except for the order) representation as a product of primes.

Congruences

We begin with

Deﬁnition B.7 Given integers a,b and m where m>0, the expression

a ≡ b(mod m) (B.1) is called a congruence, and stands for m | a −b. The number m is called the modulus, and b is called a residue of a modulo m.

This is equivalent to: a and b have the same remainder after division by m. Some- times we write a ≡ b(m) for (B.1). This was introduced by Gauss in 1801, and is one of the origins for the notions of coset and factor group. 286 Appendices A to E

Note that the congruence (B.1) is an equivalence relation on Z, see Deﬁni- tion A.1; some elementary properties are given in the problem section. Also Theo- rem B.3 can be rewritten as

Theorem B.8 Suppose a,b,m ∈ Z and m>0. The congruence

ax ≡ b(mod m) has (a, m) solutions x with 0 ≤ x

The inverse operation of the group (Z/pZ)∗ is given by this result. If m = p and 0 ≤ a

Theorem B.9 (Chinese Remainder Theorem) Given m1,...,mk ∈ Z, coprime in pairs, and c1,...,ck ∈ Z, the congruences

x ≡ ci (mod mi), i = 1,...,k, have a common solution x which is unique modulo the product m1 ···mk.

The second result, which had a formative influence on the early development of group theory and was first discovered in the seventeenth century by Fermat (and probably first proved by Euler in the eighteenth century), is

Theorem B.10 (Fermat’s Theorem) If p is prime and a ∈ Z, then ap ≡ a(mod p).

The first proofs of this result used the multinomial theorem, but nowadays a more group-theoretic proof is used; see the next result and Problem B.5. Euler extended this result to all positive moduli, to state it we need first to define a useful number-theoretic function which is named after him.

Deﬁnition B.11 If n>1, φ(n) denotes the number of integers a satisfying the conditions: 1 ≤ a

The function φ(n) is called the Euler function, it is the order of the group (Z/nZ)∗. Using Unique Factorisation in Z (Theorem B.6) it can be shown that if n has the = s1 ··· sk prime factorisation n p1 pk , then

k si −1 φ(n)= p (pi − 1). i=1 i For example, if p is prime, then φ(p)= p − 1 as every positive integer less than p is coprime to p. Note also that φ is multiplicative, that is, φ(n)φ(m)= φ(mn) when (m, n) = 1. Using this function Euler extended Fermat’s Theorem to B Number Theory 287

Theorem B.12 (Euler’s Theorem) If (a, m) = 1, then aφ(m) ≡ 1 (mod m).

Problem 2.8 when it is applied to the group Z/mZ provides a proof of this result, another is given in Problem B.5. The last congruence result is

Theorem B.13 Suppose f is a polynomial deﬁned over the ﬁeld Z/pZ, and has degree n. The polynomial congruence f(x)≡ 0 (mod p) has at most n roots in Z/pZ.

This is proved by induction on n; use Theorem B.8 when n = 1, and consider possible linear factors of f in the inductive step. This result can also be proved purely algebraically, see Problem 7.5(iii).

Primitive Roots

The basic question here is: In which cases, if any, is the group (Z/mZ)∗ cyclic? The answer is useful in the theories of cyclic groups and automorphisms (Section 4.4).

Deﬁnition B.14 (i) Given integers c and m with (c, m) = 1 and m>1, the order of c modulo m,ordm(c), is the least positive integer t satisfying the congruence ct ≡ 1 (mod m).

(ii) The integer c is called a primitive root modulo m if ordm(c) = φ(m).

By Theorem B.12,ordm(c) always exists and is not greater than φ(m); in fact, it is a divisor of φ(m).Alsoordm(c) is the order of the cyclic group generated by c in (Z/mZ)∗. The following result was ﬁrst proved by Gauss.

Theorem B.15 Primitive roots exist for each prime modulus p.

This is derived by establishing a stronger result: p has φ(p − 1) primitive roots modulo p, and one proof of this (which is very elegant) uses the Möbius function μ; see Chapter 5 in Rose (1999). There are a number of unsolved problems concerning primitive roots, for example, for large primes p only weak estimates are known for the least positive primitive root of p. Not all integers have primitive roots, for example, 8 does not have one because φ(8) = 4 and all odd integers have order 2 modulo 8. Using these results we have

Theorem B.16 Suppose p is an odd prime and let r>0. The only integers larger than 1 which have primitive roots are 2, 4, pr and 2pr .

This result follows from Theorem B.15 (Problem B.8), and it implies that the group (Z/mZ)∗ is cyclic when m = 2, 4, pr or 2pr . In the remaining cases, it can be shown that (Z/mZ)∗ is a direct product of cyclic groups. For m = 2r with r>2, Z Z ∗ × we have ( /m ) C2 C2r−2 , and the remaining cases use the Fundamental Theorem of Abelian Groups (Theorem 7.12). 288 Appendices A to E

B.2 Problems

= r1 ··· rk = s1 ··· sk Problem B.2 If m p1 pk and n p1 pk where some of the ri or sj may be zero, then

k k (m, n) = pmin(ri ,si ) and LCM(m, n) = pmax(ri ,si ). i=1 i i=1 i

Problem B.3 If a and b are positive with (a, b) = 1, show that the equation ax + by = n is soluble in non-negative integers x and y if n ≥ (a − 1)(b − 1).

Problem B.4 Show that for ﬁxed r the following equation only has a ﬁnite number of positive integer solutions {n1,...,nr }

1 1 1 = +···+ . n1 nr

This is used in Problem 5.23.

Problem B.5 Prove Theorem B.12 using the same method as for Theorem 2.8.

Problem B.6 Prove the following congruence properties. (i) If a ≡ b(mod m) and c ≡ d(mod m), then a + c ≡ b + d(mod m), ac ≡ bd (mod m), and a ≡ b(mod m) if m | m. (ii) If a ≡ b(mod m), then (a, m) = (b, m), ar ≡ br (mod mr), and at ≡ bt (mod m) when t ≥ 0. = ··· = ∗ (iii) Prove Theorem B.9 as follows: let m m1 mk, mi m/mi and mi satisfy ∗ ≡ = = ∗ +···+ ∗ mimi 1 (mod mi) for i 1,...,k, and set x c1m1m1 ckmkmk .

Problem B.7 Let the number-theoretic function χ be given by: χ(1) = χ(2) = 1, χ(4) = 2, χ(2n+3) = φ(2n+3)/2, χ(pn) = φ(pn), p an odd prime, and if m = r1 ··· rk = r1 rk p1 pk then χ(m) LCM(χ(p1 ),...,χ(pk )). Show that

aχ(m) ≡ 1 (mod m) if (a, m) = 1. C Data on Groups of Order 24 289

Problem B.8 (i) Using the Binomial Theorem, show that if r>1 and p is an odd prime, then

r−2 − (1 + ap)p ≡ 1 + apr 1 (mod pr ).

(ii) Prove that if p a, then 1 + ap has order pr−1 modulo pr . (iii) Use (ii) to show that if p is an odd prime and r>0, then pr has a primitive root. (Hint. Use Theorem B.15 and show that if c is a primitive root then cp−1 ≡ 1 (mod p2), if this fails for c try c + p.) (iv) Using Problem B.7 and (iii) prove Theorem B.16.

******

Appendix C: Data on Groups of Order 24

Data on groups of order 24 are presented in the four tables below, some notes concerning this data are given on pages 292 and 293.

Table C.1 Data on groups of order 24 GROUP No. of No. of No. of No. of No. of No. of Sylow elts of conj. cls subgps subgp max. normal 2-subgps order 2 of elts conj. cls subgps subgps

C24 1248 8 2 8C8

C12 × C2 3241616416C4 × C2 × 2 3 C6 C2 7243232832C2 D3 × C4 7122616611C4 × C2 by 3

D4 × C3 5152016412D4

Q2 × C3 1151212412Q2 × 3 D6 C2 15 12 54 32 10 21 C2 by 3 Q3 × C2 3122216613C4 × C2 by 3 × 3 ∗ A4 C2 7 8 26 12 6 6 (C2 ) D12 13 9 34 16 6 9 D4 by 3

Q6 1 9 18 12 6 9 Q2 by 3

S4 9 5 30 11 8 4 D4 by 3

F3,8 11210847C8 by 3 ∗ SL2(3) 17157 54(Q2)

E 9 9 30 16 6 9 D4 by 3 290 Appendices A to E

Table C.2 Data on groups of order 24 GROUP No. of Subgroups Maximal Centre Class Derived Sylow of order subgps No. subgp 3-subgps 12 of order 6 (ﬁrst)

C24 1 C12 Ð C24 24 e

C12 × C2 1 C12 by 2, C6 × C2 Ð C12 × C2 24 e × 2 × 2 × 2 C6 C2 1 C3 C2 by 7 Ð C6 C2 24 e D3 × C4 1 C12,D6,Q3 Ð C4 12 C3 ∗ D4 × C3 1 C12,C6 × C2 by 2 Ð C6 15 (C2) ∗ Q2 × C3 1 C12 by 3 Ð C6 15 (C2) × × 2 D6 C2 1 C C2,D6 by 6 Ð C2 12 C3 × × 2 Q3 C2 1 Q3 by 2, C6 C2 Ð C2 12 C3 × 2 A4 C2 4 A4 C6 by 4 C2 8 C2 ∗ D12 1 C12,D6 by 2 Ð (C2) 9 C6 ∗ Q6 1 C12,Q3 by 2 Ð (C2) 9 C6

S4 4 A4 D3 by 4 e 5 A4 ∗ F3,8 1 C12 Ð (C4) 12 C3 ∗ ∗ SL2(3) 4Ð C6 by 4 (C2) 7 (Q2) ∗ E 1 C6 × C2,D6,Q3 Ð (C2) 9 C6

Table C.3 Data on groups of order 24 GROUP Frattini Fitting Sockel Largest Smallest Is subgp. subgp. See (17) factor symmetric nil- group(s) supergp. potent

C24 C4 ‘G’ C6 C12 S11 Yes

C12 × C2 C2 ‘G’ C6 × C2 C12 by 2,C6 × C2 S9 Yes × 2 × C6 C2 e ‘G’‘G’ C6 C2 by 7 S9 Yes D3 × C4 C2 C12 C6 D6 S7 No ∗ D4 × C3 (C2) ‘G’ C6 C6 × C2 S7 Yes ∗ Q2 × C3 (C2) ‘G’ C6 C6 × C2 S11 Yes

D6 × C2 e C6 × C2 C6 × C2 D6 by 3 S7 No

Q3 × C2 C2 C6 × C2 C6 × C2 Q3 by 2, D6 S9 No × 3 ∗ 3 ∗ A4 C2 e (C2 ) (C2 ) A4 S6 No ∗ D12 (C2) C12 C6 D6 S7 No ∗ Q6 (C2) C12 C6 D6 S24 No 2 ∗ 2 ∗ S4 e (C2 ) (C2 ) D3 (order 6) S4 No ∗ F3,8 (C4) C12 C6 Q3 S11 No ∗ ∗ SL2(3)(C2) (Q2) C2 A4 S8 No ∗ E(C2) C6 × C2 C6 D6 S7 No C Data on Groups of Order 24 291

Table C.4 Data on groups of order 24 GROUP Degree Autom. Outer Expo- Abelian 2-Euler pattern group autom. nent invar- count See (18) order group iants See (19)

24 3 C24 1, ...,18C2 24 3, 8 6144 24 C12 × C2 1, ...,116D4 × C2 12 2, 3, 4 768 × 2 24 × C6 C2 1, ...,1 336 GL3(2) C2 6 2,2,2,3 0 8 D3 × C4 1, ...,1, 2, 2, 2, 224 C2 × C2 12 2, 4 576 12 D4 × C3 1, ...,1, 2, 2, 216 C2 × C2 12 2, 2, 3 768 12 Q2 × C3 1, ...,1, 2, 2, 248 A4 12 2, 2, 3 768 8 D6 × C2 1, ...,1, 2, 2, 2, 2 144 S4 62,2,30 5 Q3 × C2 1, 1, 1, 1, 2, ...,248 C4 × C2 12 2, 4 576 6 A4 × C2 1, ...,1, 3, 324C2 6 2, 3 288 5 D12 1, 1, 1, 1, 2, ...,248 C2 × C2 12 2, 2 576 5 Q6 1, 1, 1, 1, 2, ...,248 C2 × C2 12 2, 2 576

S4 1, 1, 2, 3, 324e 12 2 216 8 F3,8 1, ...,1, 2, 2, 2, 224 C2 × C2 24 8 4608

SL2(3) 1, 1, 1, 2, 2, 2, 324 C2 12 3 384 5 E 1, 1, 1, 1, 2, ...,224 C2 12 2, 2 576

Notes on Tables C.1 to C.4

(1) We write H k as a short-hand for H × H ×···×H with k factors.

(2) ‘H by k’ stands for ‘k isomorphic copies of the subgroup H ’.

(3) If ‘(H )∗’ occurs twice or more in a row(s), this means that the corresponding subgroups are identical. For example, the derived and Frattini subgroups of D4 × C3 are identical.

(4) The group Q2 × C3 is the only group in the tables that is Hamiltonian: It is not Abelian and all of its subgroups are normal; see Problem 7.13 and Note (20).

(5) The tabulated maximal subgroups have order 12 (except for SL2(3)), or order 8 (all Sylow 2-subgroups are maximal), or in three cases (A4 × C2,S4 and SL2(3)) they have order 6. 292 Appendices A to E

(6) All largest factor groups have order 12 except for S4 where the order is 6. This reﬂects the fact that all groups of order 24 except S4 have at least one normal subgroup of order 2.

(7) The centre is deﬁned on page 32, and the derived subgroup on page 37, only S4 and SL2(3) have non-neutral second derived subgroups, they are isomorphic to C2 × C2 for S4 and C2 for SL2(3).

(8) Nilpotency and supersolubility are deﬁned on page 212 and page 227,respectively. The two non-Abelian nilpotent groups listed above (D4 × C3 and Q2 × C3) both have nilpotency class 2. All groups in the table are soluble, and all but A4 × C2, S4 and SL2(3) are supersoluble; see Problem 10.26.

(9) The exponent (Table C.4) of a group G is the smallest positive integer n such that gn = e for all g ∈ G (Deﬁnition 2.19(iv)).

(10) The Abelian invariants (Table 4) of G are deﬁned as the orders of the elements of a minimal set of generators of ‘G made Abelian’, that is, of G factored by its derived subgroup G.

(11) The Frattini subgroup (G) of G is the intersection of the maximal subgroups of G, it is a normal subgroup of G, see page 217.

(12) The Fitting subgroup F(G) of G is the (unique) maximal, nilpotent, normal subgroup of G; see page 221.IfG is nilpotent then F(G) = G.

(13) The groups in the tables with the maximum number of elements of the speci- ﬁed order are as follows: D6 × C2 has 15 elements of order 2, A4 × C2,S4 and × 2 SL2(3) have 8 elements of order 3, Q6 has 14 elements of order 4, C6 C2 has 14 elements of order 6, F3,8 has 12 elements of order 8, Q2 × C3 has 12 elements of order 12, and C24 is the only group with elements (8 in all) of order 24.

(14) Cauchy’s Theorem states that all the groups in the tables possess elements of order 2 and 3. For all other divisors n of 24, there is at least one group in the × 2 × table which does not have an element of order n. The groups C6 C2 ,D6 C2 and A4 × C2 do not have elements of order 4, S4 does not have an element of order 6, all groups in the table except C24 and F3,8 do not have elements of × 2 × × × order 8, C6 C2 ,D6 C2,Q3 C2,A4 C2,S4, SL2(3) and E do not have elements of order 12, and no group except C24 has elements of order 24.

(15) By Corollary 5.27, the order of the inner automorphism group times the order of the centre equals 24, so o(Inn G) ranges from 1 (for Abelian groups) to 24 (when G = S4). D Numbers of Groups with Order up to 520 293

(16) With one exception all groups in these tables are reverse Lagrange, that is, they possess subgroups of all orders dividing 24. The exception is SL2(3) which has no subgroup of order 12.

(17) The Sockel of a group G is the product of all minimal normal subgroups of the group G, see Theorem 11.6. It is normal in G.TheAbelian,(non-Abelian, Sockel) is the product of all Abelian, (non-Abelian, respectively), minimal normal subgroups of G, see Theorem 11.11. The sockel is the product of the Abelian and non-Abelian sockels. A number of properties have been established, for example, if G is a ﬁnite group and (G) =e, then F(G) equals the Abelian Sockel. For further details, see Scott (1964), pages 168 to 170.

(18) The degree pattern of a group G is a set of positive integers (r1,...,rm) where ri is the degree of the ith linear representation of G (that is, the dimension of m 2 = its underlying vector space). Note that i=1 ri o(G). This aspect of G is mainly concerned with the character theory of G; see Huppert (1998), and Web Section 14.3.

(19) The 2-Euler count is the number of pairs of (possibly not distinct) elements of a group G that generate G.

(20) All groups except C24 in the tables can be represented as direct or semi-direct products with two factors; see Problem 8.3. Apart from those listed in this chapter we have D12 C2 C12, Q6 Q2 C3, and F3,8 C8 C3.Note also that Q2 × C3 can also be treated as a semi-dihedral group, see page 131.

(21) Most of the data in these tables was computed using the computer package GAP. The numbers (labels) given to the groups by the GAP program ‘Small- Groups’ are: 2, 9, 15, 5, 10, 11, 14, 7, 13, 6, 4, 12, 1, 3, and 8, using the same order of the groups as in the tables above—so for example, the GAP SmallGroup(24, 1) is the thirteenth group in our tables, that is, F3,8.

******

Appendix D: Numbers of Groups with Order up to 520

The table apposite gives the number of (isomorphism classes of) groups of each order between 1 and 520 except 512 which is a staggering 10494213; see Besche et al. (2001). The entries 1∗ indicate that the number of groups of this order is 1 even though the order is not prime, there are 73 in Table D.1, see page 117. The data in this table were given by the programs GAP and Magma. 294 Appendices A to E

Table D.1 Number of Groups with Order up to 520 + 0 40 80 120 160 200 240 280 320 360 400 440 480 1 111521∗ 2111∗ 21131∗ 2 1622552 5 44 26 42 3 1111∗ 12 6711 31∗ 12 4 2 4 15 4 5 12 5 4 176 11 5 18 12 5 121∗ 5222221∗ 16 1∗ 1∗ 6 2221622 4 42 66 2261 7 111∗ 1121∗ 1∗ 211∗ 1∗ 1 8 5 52 12 2328 57 51 12 1045 15 42 46 1396 14 9 2212 21∗ 1∗ 21∗ 21 12 10 2 5 10 4 4 12 15 4 12 4 6 34 10 11 11∗ 1∗ 1511211∗ 1∗ 1∗ 1 12 55410454654154512 13 1121∗ 11∗ 21511∗ 21∗ 14 2152242223241024 15 1∗ 21∗ 521∗ 1∗ 1∗ 1∗ 71∗ 1∗ 4 16 14 13 231 15 42 177 56092 14 228 12 235 54 42 17 12111∗ 1∗ 1511∗ 212 18 5254 22 6 25 60424 19 1121 12 1∗ 1∗ 1∗ 11 51 20 5131611371515491511411156 21 2111∗ 11∗ 221∗ 21 11∗ 22 2242 46 2 21822122 23 1411∗ 21 11∗ 51211 24 15 267 14 197 12 197 39 42 12 20169 14 51 202 25 21∗ 21∗ 1∗ 61∗ 21∗ 22 42 26 2422 62 410224 26 27 51161∗ 11∗ 11 41∗ 16 28 4 5 45 5 4 15 4 9 12 5 4 55 4 29 11∗ 11131 1 21 121∗ 1 30 44613443061012448 31 1121 12 1 1141∗ 121∗ 32 51 50 43 12 1543 14 54 61 195 44 775 12 ∗∗∗ 33 1∗ 112 11 5 11 1∗ 11∗ 15 34 2264 2162 24 24 62 35 1∗ 31∗ 221∗ 4421∗ 1∗ 21∗ 36 14 4 5 18 12 4 10 4 5 30 5 11 15 37 11∗ 41 12 1 12 11∗ 21∗ 38 2622104 2 42 26 24 39 211∗ 1∗ 11 41∗ 15111∗ 40 14 52 47 238 52 208 40 1640 162 221 51 1213 49 E Representations of L2(q) with Order < 10000 295

Appendix E: Representations of L2(q) with Order < 10000

In this appendix, we list some representations for the linear groups L2(q), see Sec- tion 12.2. All representations are illustrative examples and are in no way unique. A presentation for L2(p), p a prime, was given in Chapter 12. In the ﬁrst list, we give some presentations for the remaining cases using small order generators. They were taken from the ATLAS (1985) where further details can be found. 7 2 3 3 2 5 2 L2(8) a,b | a = (a b) = (a b) = (ab ) = e , 5 5 2 4 4 L2(9) a,b | a = b = (ab) = (a b) = e A6, 15 2 3 3 2 9 2 8 2 2 L2(16) a,b | a = (a b) = (a b) = (ab ) = (a b ) = e , 5 12 2 2 3 4 3 10 2 11 2 11 2 L2(25) a,b,c | a = b = c = (bc) = (ac) = (b abac) = e,b ab = b a ba = ab a b , 3 13 2 2 3 10 3 12 12 L2(27) a,b,c | a = b = c = (bc) = (ac) = e,b ab = ab ab = b aba . Secondly, each linear group in our chosen range can be given a permutation representation, that is, L2(q) is isomorphic to a subgroup of At for some suitable chosen t depending on q. In the list below which was constructed using the computer program GAP, t = q + 1, but in some cases smaller values of t are possible (for q = 5, 7, 9). The case q = 11 is discussed in Problem 12.8. These facts were ﬁrst noted by Galois in his famous letter to a colleague written the night before he was mortally wounded in a duel; see Huppert (1967), page 214. L2(4) (3, 4, 5), (1, 2, 3) A5, L2(5) (1, 2, 3, 4, 5), (3, 4, 5) A5, L2(7) (1, 2)(3, 4), (1, 3, 5)(2, 7, 6) A7, L2(8) (3, 9, 7, 4, 8, 5, 6), (1, 2, 3)(4, 5, 6)(7, 8, 9) , L2(9) (1, 2, 3, 4, 5), (4, 5, 6) A6, L2(11) (1, 3)(2, 5)(7, 10)(9, 11), (2, 6, 8)(3, 9, 5)(4, 11, 10) M11, L2(13) (3, 13, 11, 9, 7, 5)(4, 14, 12, 10, 8, 6), (1, 2, 9)(3, 8, 10)(4, 5, 12)(6, 13, 14) , L2(16) (3, 16, 14, 13, 11, 4, 8, 6, 17, 15, 5, 12, 10, 9, 7),

(1, 2, 3)(6, 14, 8)(7, 11, 10)(9, 17, 15)(12, 16, 13) , L2(17) (3, 17, 15, 13, 11, 9, 7, 5)(4, 18, 16, 14, 12, 10, 8, 6),

(1, 2, 11)(3, 13, 9)(4, 15, 6)(5, 8, 12)(7, 18, 16)(10, 14, 17) , L2(19) (3, 19, 17, 15, 13, 11, 9, 7, 5)(4, 20, 18, 16, 14, 12, 10, 8, 6),

(1, 2, 12)(3, 11, 13)(4, 17, 6)(5, 14, 18)(7, 20, 18)(10, 19, 16) , L2(23) (3, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5)(4, 24, 22, 20, 18, 16, 14, 12, 10, 8, 6),

(1, 2, 14)(3, 12, 16)(4, 18, 9)(5, 20, 6)(7, 13, 11)(8, 23, 22)(10, 24, 19)(15, 21, 17) , L2(25) (3, 25, 23, 6, 20, 18, 5, 15, 13, 4, 10, 8)(7, 26, 24, 22, 21, 19, 17, 16, 14, 12, 11, 9),

(1, 2, 5)(3, 4, 6)(7, 18, 14)(8, 22, 9)(10, 17, 12)(11, 25, 24)(15, 26, 19)(16, 23, 21) , L2(27) (3, 27, 25, 23, 21, 19, 17, 16, 14, 12, 10, 8, 6)(4, 15, 13, 11, 9, 7, 5, 28, 26, 24, 22, 20, 18),

(1, 2, 4)(5, 8, 24)(6, 21, 10)(7, 16, 15)(9, 25, 28)(11, 13, 14)(12, 27, 23)(17, 26, 28) . Bibliography

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Three notation indexes are given below: 1—Symbol index; 2—Index of names of classes of groups; and 3—Index of names of individual groups.

1—Symbol Index

,11 H ≤ G, H

T1,12 g,26 (G, ), G, H, J, K,13 o(g),26 X, 13, 25 XY,gH,Hg,27 e, 13, 24 [G : H ],29 a,b,c,d,g,h,j,k,13 ,30 ab,e,g−1,15 K,30 ∨ o(G),o(X),16 H J ,31 Z, 18, 21 H,J ,31 , 17, 69 Z(G), 32, 102 ∗ ∗ ∗ [g,h], [G, H ], 37, 210 Q, R, C, Q , R , C ,18 G ,37 Q+, R+,18 core(H ), 39, 101 F ∗ (non-zero elements of F ), 18 H ∗,39 a ≡ b(mod m), a ≡ b(m), 18, 285 12... n Z/mZ, 18, 74 ,42 a1 a2 ... an (Z/pZ)∗,18 (a1,...,an) (cycle), 44 T2,Tn,18 k-cycle, 44 GLn(F ), 19, 53 sgn(σ ),47 SLn(F ), 19, 53 GLn(q), SLn(q),53 UTn(F ), 19, 54 Ln(q), 54, 255 Dn, 20, 58 A | R, g1,...| x1 = e,...,58 SX,Sn, 20, 49 Z,58 An, 20, 50 Qn,59 Cn, 22, 80 ITn(F ), IZTn,r (F ),63

H.E. Rose, A Course on Finite Groups, 301 Universitext, DOI 10.1007/978-1-84882-889-6, © Springer-Verlag London Limited 2009 302 Notation Index

n φ|H ,70 Sz(2 ), 222 2 im φ, 70, 280 F4(2) , 222 ker φ,71 Gal(F ), 229, 230 G/K,73 G(n), 234 (isomorphic to subgroup), 75 π,π , 236, 285 Aut G,Aut(G),81 S(r,s,t), 251 Inn G,Inn(G),81 Aut(S(r,s,t)), 253 Out G,Out(G),82 Fq , 254 char, 89 PSLn(q), 255 \g,91ff Eij (r), 255 \ (action), 92 GUn(q), SUn(q), Un(Q), 265 O{x}, OG{x},94 M11,M12,..., 267, 268 stabG(x),95 J1,J2, 267 ﬁx(G, X),98 ∈, 277 C{x}, CG{x}, 102 =, 277 CG(x), 102 ∅, 277 h(G), 103 ⊆, ⊂, 277 CG(X), 104 ∩, 278 NG(H ), 105 ∪, 278 v(n), 117 \ (complement), 278 Fm,n, Fn,m,r , 130 {x ∈ X : P(x)}, 278 ES1(p), ES2(p), 132 {x1,...,xn}, 278 ×, 139 (x1,...,xn), 278 G1 ×···×Gn, 140 ×, 278 part(n), 151 Xn, 278 , 151 φ : X → Y , 279 K : A, 152 Zφ, 280, 281 wr, , 156 Zφ−1, 280, 281 Hol(G), 162 ι, 281 PGLn(q), 170 ◦, 281 Gr,s,t , 170 (partial order), 282 E, 177 , 282 , 206, 224 a | b, 284 Di (G), Di , 210 a b, 284 Zi (G), Zi , 210 (a, b), 284 (G), 217 GCD, LCM, 284 F(G), 221 ≡, mod, 285 Op(G), Oπ (G), 221, 237 φ(n), 286 ∗ F (G), 222 ordm(c), 287 Notation Index 303

2—Notation for Classes of Groups

In the first table below, we list the primary definitions and notations for the infinite classes of groups discussed in this book. Note that in all cases the subscript n is used to indicate the nth group in the class, and so is only indirectly related to the order of the group. The column called ‘Primary definition’ gives the group’s ‘main’ representation. The symbol F denotes an arbitrary field, if F is finite then the symbol F is replaced by q the order of the field.

Group English Primary Page symbol name deﬁnition

An Alternating Even permutations on n symbols 50

Cn Z/nZ Cyclic Integers with addition modulo n 58, 79

Dn Dihedral Symmetries of regular n-gon 58 r s −1 t Fr,s or Frobenius or a,b | a = b = e,b ab = a 131 s Fr,s,t metacyclic where t ≡ 1 (mod r)

GLn(F ) General linear Non-singular n × n matrices 53 over F

ITn(F ) Uni-upper Subgroup of UTn(F ) with each 63 triangular diagonal element 1

IZTn,r (F ) Ð Subgroup of ITn(F ) with r 63 superdiagonals zero

Ln(F ) or Linear or SLn(F ) factored by its centre 255 PSLn(F ) Projective special linear n 2 2 Qn Dicyclic or a,b | a = b = (ab) 59 generalised quaternion

Sn Symmetric All permutations on n symbols 49

SLn(F ) Special linear Subgroup of GLn(F ) of matrices A 53 with det A = 1 Sz(2n) Suzuki See Web Section 14.3 Ð n Tn C2 n copies of ‘2’ Ð 18 and 151

Un(F ) Unitary See page 265 265

UTn(F ) Upper Subgroup of GLn(F ) of matrices 54 triangular with zeros below main diagonal (Z/nZ)∗ Ð {a ∈ Z : 0

3—Notation for Individual Groups

In the second table below, we list some notation for individual groups, that is, groups that do not form a part of an inﬁnite family.

Group English name Operation Page

C Complex numbers Addition 18 C∗ Non-zero complex Multiplication 18 numbers

Cp∞ Inﬁnite group with [Abelian] 132 only ﬁnite subgroups E Exceptional See Section 7.3 177

ESi for Extra special See Problems 3.18 and 5.5 132 i = 1, 2

J1,J2 Janko Ð 267

M11,M12 Mathieu See Section 11.4 268 4 2 2 −1 3 Q2 Quaternion a,b | a = e,a = b ,b ab = a 118 Q Rational numbers Addition 18 Q∗ Non-zero rational Multiplication 18 numbers Q+ Positive rational Multiplication 18 numbers R Real numbers Addition 18 R∗ Non-zero real Multiplication 18 numbers R+ Positive real Multiplication 18 Z Integers, or inﬁnite Addition 18, 58 cyclic group Index

2-Euler count, 293 B 4-group, 19 Basis Theorem, 147 Belong (∈), 277 Bertrand’s Postulate, 52 A Bijection, 280 Abel, N, 12 Binary (relation), 279 Abelian, 2, 12 tetrahedral (group), 176 group, ﬁnite, 146ff Block (of a Steiner system), 251 elementary, 88 Bruhat Decomposition Theorem, 55 invariant, 292 Burnside, W. 27, 74, 130, 220 maximal, 109 Normal Complement Theorem, 130 r s Action, 91ff p q -theorem, 231, 239 conjugate element, 101 subgroup, 105 C coset, 100 Cancellation, 15 intransitive, 94 Canonical (rational form), 262 natural, 93 Carbon60, 23 permutation, 94 Cardano, G. 243 restricted, 98 Cardinality, 16, 280 right multiplication, 93 Cartesian (product), 278 transitive, 94 Cauchy, A. 42, 114 Algorithm, division, 284 Theorem, 114 Cayley, A. 1, 42, 48, 72 Euclidean, 284 number, 119 Alphabet, 21, 56 Theorem, 42, 72 Alternating (group), 20, 50, 61 (A ), 158 4 Central, 109 Antisymmetric, 282 series, 210 Argument, 280 Centraliser, 101ff, 104, 108 Frattini, 128 Centre, 32, 87, 102 Arithmetic, modular, 18 higher, 210 Associative, 2, 11 Centreless, 32 Automiser, 106 CFSG, 1 Automorphism, 69 Characteristic (of a ﬁeld), 254 group, 81ff subgroup, 30, 89 inner, 82 Characteristically simple, 161 outer, 82 Chief series, 207 of a Steiner system, 252 ChordÐtangent process, 22

H.E. Rose, A Course on Finite Groups, 305 Universitext, DOI 10.1007/978-1-84882-889-6, © Springer-Verlag London Limited 2009 306 Index

Class Equations, 103 Descartes, R. 243, 278 equivalence, 279 Dicyclic (group), 59 number 103 generalised, 205 Closure,2,11 Dihedral (group), 4, 58, 64 Cocycle Identity, 198 Dining Club, 270 Codomain, 280 Direct product, 139, 140 Cohomology theory, 187 external, 140 second group, 197 internal, 140, 141 Collection, 277 Disjoint (sets), 278 Combinatorial group theory, 55 Divide, 284 Commutative, 12 Division algebra, 119 Commutator, 37 algorithm, 284 identities, 37 Domain, 280 subgroup, 37 Double cosets, 40 Companion matrix, 262 Dyck, W. von, 55 Complement (set), 278 Frobenius, i E subgroup, 151 Element, 277 p-, 239 neutral, 2, 4, 12 Composite, 285 Elementary Abelian group, 27, 88, 162 Composition factor, 188 Empty set, 277 (of maps), 281 word, 21, 56 series, 188, 190 Endomorphism, 69 Concatenation, 21, 56 Equation, Class, 103 reduced, 56 Equality (function), 280 Congruence, 18, 285 (group), 16 Conjugacy class, 30, 102 (set), 277 action, 102, 105 Equivalence class, 279 Conjugate element, 30 relation, 279 subgroup, 38, 105 Equivalent (series), 191 Coprime, 285 Euclid, 284, 285 Core, 39, 100 Euclidean Algorithm, 284 Correspondence Theorem, 78 Euler’s function, 288 Coset (left and right), 27 Theorem, 288 action, 100 Even (permutation), 47ff double, 40 Exceptional group, 177ff enlargement, 87 Exponent (of a group), 26, 291 enumeration, 58 (rules), 16 multiplication, product, 73 Extension, 151, 196, Coxeter group, 64, 269 cyclic, 196, 203ff Cycle (permutation), 44 problem, 196ff of length k,44 split, 151 Cyclic, 26 External (direct product), 140 extension, 196, 203ff (semi-direct product), 151 group, 58, 80 Extra-special (group), 132

D F Degree pattern, 293 Factor (of a direct product), 140 Dekekind, R. 161 group, 72ff Modular Law, 38 pair, 197, 198 De Morgan’s Law, 282 (of a series), 188 Derived length, 234 Factorisable (group), 152 series, 234 Factorisation, unique, 285 subgroup, 37, 85 Fano plane, 253 Index 307

Feit, W. 231 presentation, 41, 55ff Fermat,P.286 product, 18, 140ff theorem, 286 simple, 33 Ferrari, L. 243 soluble, 5, 230ff del Ferro, S. 243 supersoluble, 227 Field, finite, 254 Group examples Galois, 53 alternating, 20, 50, 158 splitting, 230 Chevalley, 249 Finite field, 254 classical, 249 geometry, 253 Coxeter, 64 Fitting, H. 221 cyclic, 58, 80 subgroup, 222 dicyclic, 59 Fixed set, 98 dihedral, 4, 20, 58, 159 Fontana (Tartaglia), N. 243 elementary Abelian, 27, 88 Frattini, G. 217 Exceptional, 177ff Argument, 128 extra-special, 132 subgroup, 217ff Frobenius, 62, 130, 184, 241 Free (group), 21, 57 Galois, 213 Friendly giant, 20 general linear, 13, 19, 52 Frobenius complement, i unitary, 265 group, 62, 131, 184, 241 generalised dicyclic, 205 Fully invariant subgroup, 30 Hessian, 265 Function, 279 infinite cyclic, 21, 58 n-variable, 280 Janko, 267 Fundamental theorem of Abelian groups, 146ff Klein, 19 linear, 52, 249, 254ff G Mathieu, 250, 264ff Galois, é. 1, 29, 30, 229, 242, 246, 261, 295 metacyclic, 62, 130, 184, 241 field, 53, 254 neutral, 13 Gauss, C.F. 18, 28, 83, 285, 287 octahedral, 171 General linear group, 13, 19, 53 projective general linear, 170 Generalised associativity law, 34 special linear, 255 dicyclic group, 205 quaternion, 59, 119, 159 Fitting subgroup, 223 second cohomology, 197 Generate (a group), 13, 25 semi-dihedral, 131 Generating set, 13, 25 special linear, 19, 52, 172 Generator, 21, 26, 58 unitary, 265 Greatest common divisor (GCD), 284 sporadic, 249, 267 Group, 12 Suzuki, i, 223 Group attributes symmetric, 13, 49, 165 automorphism, 81ff tetrahedral (A4), 61 cyclic Sylow subgroups, 131 binary, 176 factor, 72ff Tits, 223 free, 57 unitary, 265 fundamental (of a topological space), 23 G-set, 92 Hamiltonian, 161 maximal normal, 189 H nilpotent, 129, 144, 210, 212 Hall, P. 1, 236 of odd order, 231 π-subgroups, 76, 236, 246, 247 order 8, 118 ÐWitt Identity, 37 order 12, 156 Hamilton, W. 119 outer automorphism, 82, 250 Hamiltonion (group), 119, 161, 292 quotient, 72 Higher centre, 210 perfect, 86 commutator subgroup, 210 308 Index

Hölder, O. 131, 190, 196 Left coset, 27 Holomorph, 162 inverse, 13 Homomorphism, 67ff neutral element, 13 equation, 67 Length, of a cycle, 43, 44 natural, 76 of a series, 188 Hypercentre, 211 derived, 234 Letter, 56 Linear group, 52, 254 I order, 282 Idempotent, 14 List, 56 Identity (of a group), 12 Lower central series, 211 Cocycle, 198 HallÐWitt, 37 map, 281 M Image, 280 Map, mapping, 279 identity, 281 Indecomposable, 143 natural, 67 Index (of subgroup), 29 structure-preserving, 67 Inﬁnite (cyclic group), 58 Mathieu, É.L. 268 Injective, 280 group, 20, 250, 268 Internal (direct product), 141 Matrix, non-singular, 52 (semi-direct product), 151 permutation, 62 Intersection, 278 scalar, 256 Intransitive (action), 94 upper triangular, 54 Invariant series, 188 Maximal, Abelian, 109 Abelian, 292 normal subgroup, 189 Inverse, 2 subgroup, 24 of a bijection, 281 McKay, J. 114 Involution, 4, 27, 39, 113 Member, membership, 277 Isometry, 34 Metacyclic (group), 62, 131, 184 Isomorphism, isomorphic, 17, 67ff Minimal, normal subgroup, 235 class, 17 simple group, 264 of a Steiner system, 252 Modulus, modulo, 18, 285 Theorems, 75ff arithmetic, 18 Iwasawa’s Lemma, 7, 92, 257 Monster, 20 Morphism, 67 J Multiplication, 13 Janko, Z. 267 coset, 73 Join (of subgroups), 31 scalar, 93 Jordan, C. 29, 69, 190 table, 36 –Hölder Theorem, 190ff Multiplicative, 286

K N n k-cycle, 43 -argument, 280 Natural action, 93 Kernel, 30, 71ff homomorphism, 76 Kirkman, T.P. 251 map, 67 Klein (group), 19 N/C-theorem, 106 k-transitive, 94, 268 Neutral element, 2, 4, 12 group, 13 L subgroup, 24 Lagrange, J.-L. 1, 29 Nilpotency class, 210, 212 reverse, 101 Nilpotent, 129, 144, 210ff Theorem, 30 with class r, 212 Least common multiple (LCM), 284 Noether, E. 74 Index 309

Non-generator, 217 Product, 13 Normal (subgroup), 30 cartesian, 278 closure, 39 coset, 73 conditions, 30, 71 direct, 139ff minimal, 235 group, 19 properties, 36 of permutations, 42 series, 188 semi-direct, 151ff Normaliser, 105, 109 wreath, 156 Number, Cayley, 119 Projection, 283 class, 103 Projective, General Linear Group, 170 n-variable, 280 Special Linear Group, 255 Unitary Group, 265 O Proper, refinement, 188 Octahedral (group), 170 series, 188 Octonion, 119 subgroup, 24 Odd (permutation), 47ff subset, 277 One-to-one, 280 correspondence, 280 Q Onto, 280 Quaternion, 119 Operation, 11, 13, 281 group, 59, 119, 159 Orbit, 94ff Quotient (group), 73 compared with a cycle, 94 of an element, 94 Order of an element, 26 R of a group, 16 Radical, p-, 86, 222, 237 function, 35 (type of root), 229 modulo an integer, 287 Range, 280 partial, 282 Reduced word, 21, 56 Ordered set, 278 concatenation, 56 Outer, automorphism, 82 Refinement (of a series), 188 group, 82, 250 Reflexive (order), 282 (relation), 279 P Regular, 60 Partial order, 282 Relation, 21, 57, 58, 279 system, 12 equivalence, 279 Partition, 151 Relator, 58 p-complement, 239 Representation, 4, 9, 41ff Perfect (group), 86 permutation, 97 Permutation, 12, 42ff Residue, 285 action, 94 Restricted, action, 98 even and odd, 44, 47ff,62 Restriction (of a function), 70 group, 48ff Reverse Lagrange, 101 matrix, 62 Right coset, 27 product, 42 multiplication (action), 93 representation, 97 Root, primitive, 287 p-group, 114ff Russell, B. 41 Poincaré, H. 23 Theorem, 38 S Preimage, 280 Scalar matrix, 256 Presentation, 58 multiplication, 93 of Sn,64 Schmidt, O. 246 p-radical, π-radical, 86, 133, 222 Schreier, O. 82, 193 Primitive root, 287 Conjecture, 82 Prime, prime number, 285 Refinement Theorem, 195 310 Index

Schur, I. 187 derived (commutator), 37 multiplication, 250 higher, 210 Section, 197 Frattini, 217 Self-normalising, 106 Fitting, 222 Semi-dihedral (group), 131 generalised, 223 Semidirect (product), 151ff fully invariant, 31 internal, 151 identity, 24 external, 153 Hall, 236, 247 Semigroup, 11 lattice, 32, 169, 175, 180 Semi-regular, 60 maximal, 24 Semi-simple, 196 minimal normal, 235 Series, 188ff n-derived, 234 central, 210 neutral, 24 chief, 207 normal, 30 composition, 188ff proper, 24 derived, 234 subnormal, 206 equivalent, 191 Sylow, 122 invariant, 188 trivial, 24 lower central, 211 unit, 24 normal, 188 Subnormal (series), 188 subgroup 206, 224 soluble, 230 Subset, 277 subnormal, 188 Supersoluble (group), 227 upper central, 211 Surjective, 280 Set, 277 Suzuki,M,i,52 ﬁxed, 98 group i, 223 generating 13, 25 Sylow, L. 120 ordered, 278 p-subgroup, 81, 122 underlying, 13 subgroup, 1, 120ff Sign (of a permutation), 49 properties, 133, 134 Simple (group), 33 system, 241 characteristically, 161 theorems, 120ff Single-valued, 279 Sylow 1, ..., Sylow 5, 125 Size, 16, 280 Symbol, 56 Sockel, 177, 293 Symmetric, 279 Solubility conditions, 241ff (group), 13, 48, 165 Soluble (group), 5, 116, 230 difference, 282 series, 230 Symmetry, 3 Solvable, 230 of a tetrahedron, 61 Special Linear Group, 53, 172 science of, 42 Sphere packing, 20 Split (extension), 152, 229 T Splitting ﬁeld, 230 Tarski Monster, 132 Sporadic (group) 249 Term (of a series), 188 Stabiliser, 95ff Ternary (relation), 279 Steiner, J. 251 Tetrahedron, 61 system 250ff Theorem, Basis (of Abelian Groups), 147 automorphism, 252 Brahut Decomposition, 54 isomorphism, 252 Burnside’s Normal Complement, i, 130 Structure-preserving map, 67 pr qs -, i, 231, 242, 245 Subgroup, 24 Cauchy, 114 characteristic, 31, 89 Cayley, 72 condition, 24, 25 Correspondence, 77ff conjugate, 38, 105 Euler, 286 Index 311

Fermat, 35, 286 U Fitting, 221 Underlying set, 13 Freshman’s, 79 Unimodular (matrix), 147 Fundamental, of Abelian Groups, 146ff, 162 Union (of sets), 278 Isomorphism, First, 75 Unipotent (matrix), 63 Second, 77 Unique factorisation, 285 Third, 79 Unitary group, 265 Jordan–Hölder, i, 190ff matrix, 265 Unordered (set), 278 Lagrange, 27, 29, 30 Upper central series, 211 N/C, 106 triangular matrix, 19, 54 Orbit-stabiliser, 95 Poincaré, 38 V Schreier Reﬁnement, 195 Variable, 280 SchurÐZassenhaus, i, 187 Viergruppe (4-group), 19 Sylow, i, 120ff Zassenhaus, 193 W Thompson, J. 231, 242, 246 Wedderburn, J.H.M. 143 Three Subgroup Lemma, 225 Well-deﬁned, 2, 11, (coset) 73 Tits group, 223 Well-order, 282 Total (order), 282 Wielandt, H. 119 Transfer, 68 Word, 21, 56 Transformation, 279 Wreath (product), 156 Transitive, 279 action, 94 Z order, 282 Zassenhaus, H. 130, 187, 193 Transposition (2-cycle), 44 Lemma, 193 Transvection, 55, 255 Schur (theorem), 187 Trivial (homomorphism), 69 Zero, 12